IPMAT Indore Circles PYQs

Q1:ipmat indore 2020QA › CirclesMediumSA · TITA

The shortest distance from the point

(-4,3)

to the circle

(x^2 + y^2 = 1)

is __________.

Enter your answer to attempt

The Setup: This is a coordinate geometry problem finding the shortest path from an external point to a circle's boundary. Think of the circle as a forcefield; the shortest route is always a straight line aimed dead center, and we just subtract the forcefield's radius from the total distance. Step 1: Deconstruct the circle's equation. The given equation is

x^2+y^2=1

. This perfectly matches the standard form

(x-h)^2+(y-k)^2=r^2

. Thus, the center is the origin

(0,0)

and the radius is

r=\sqrt{1}=1

. Step 2: Calculate absolute distance to center. We use the Euclidean distance formula between the point

(-4,3)

and the center

(0,0)

:

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

d=\sqrt{(-4-0)^2+(3-0)^2}

d=\sqrt{(-4)^2+(3)^2}

d=\sqrt{16+9}=\sqrt{25}=5

Step 3: Confirm the point is outside. This step is not optional - the *subtract the radius* move only works for an external point. Since

d=5>r=1

, the point sits outside the circle. (Had it been inside, the shortest distance would have been

r-d

instead.) Step 4: Find the boundary distance. The total distance to the core is

5

units. Because the circle's radius occupies

1

unit of that path, the remaining gap from the point to the edge is simply the difference:

5-1=4

. Final Answer: 4

Q2:ipmat indore 2023QA › CirclesHardSA · TITA

In the xy-plane let

A = (- 2, 0), B = (2, 0)

. Define the set S as the collection of all points C on the circle

x ^ 2 + y ^ 2 = 4

such that the area of the triangle ABC is an integer. The number of points in the set S is

Enter your answer to attempt

The Setup: Points

A=(-2,0)

and

B=(2,0)

lie on the

x

-axis. We must find all points

C

on the circle

x^2+y^2=4

such that the area of

\triangle ABC

is an integer. Step 1: Define the area of the triangle. The base of the triangle is the segment

AB

connecting

(-2,0)

and

(2,0)

. Its length is exactly

4

. Let point

C

have coordinates

(x, y)

. The height of the triangle relative to base

AB

is the absolute value of the

y

-coordinate,

|y|

.

\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times |y| = 2|y|

Step 2: Establish integer constraints for the area. We require the Area (

2|y|

) to be a positive integer

k

(a valid geometric triangle must have an area strictly greater than 0).

2|y| = k \implies |y| = \frac{k}{2}

Since

C

lies on the circle

x^2+y^2=4

, the maximum possible value for

|y|

is the radius,

2

.

0 < |y| \le 2 \implies 0 < \frac{k}{2} \le 2 \implies k \in \{1, 2, 3, 4\}

Step 3: Map the valid area states back to coordinate points. For each valid

k

, we substitute

|y| = k/2

back into the circle equation

x^2 = 4 - y^2

to count the distinct points: * If

k = 1

:

|y| = 0.5 \implies y = \pm 0.5

.

x^2 = 4 - 0.25 = 3.75

. This yields

2

values for

x

per

y

, giving

4

points. * If

k = 2

:

|y| = 1 \implies y = \pm 1

.

x^2 = 4 - 1 = 3

. This gives

4

points. * If

k = 3

:

|y| = 1.5 \implies y = \pm 1.5

.

x^2 = 4 - 2.25 = 1.75

. This gives

4

points. * If

k = 4

:

|y| = 2 \implies y = \pm 2

.

x^2 = 4 - 4 = 0

. Here

x = 0

uniquely, giving exactly

1

point for

y=2

and

1

point for

y=-2

, totaling

2

points. Step 4: Sum the valid points.

\text{Total Points} = 4 + 4 + 4 + 2 = 14

Final Answer: 14

Q3:ipmat indore 2025QA › CirclesMediumSA · TITA

A circle of radius 13 cm touches the adjacent sides AB and BC of a square ABCD at M and N, respectively. If

AB=18

cm and the circle intersects the other two sides CD and DA at P and Q, respectively, then the area, in sq. cm, of triangle PMD is

Enter your answer to attempt

The Setup: We are dropping into a coordinate geometry map for this one. We have a square bounding box and a circle spawning inside it, touching the walls. By setting a strategic origin point, we can turn this messy geometry problem into a pure algebra speedrun. I have fully audited the coordinates to ensure the hitbox interactions are completely accurate. Step 1: Setting the Origin (The Spawn Point) Let's anchor the square's corner

B

exactly at the origin

(0,0)

. Since the square's side length is

AB = 18

cm, we can lock in the corner coordinates: *

B = (0,0)

*

C = (18,0)

*

A = (0,18)

*

D = (18,18)

Step 2: Defining the Circle's Hitbox The circle has a radius

r = 13

cm. It touches side

AB

(which sits on the y-axis) and side

BC

(which sits on the x-axis). Because it is perfectly tangent to both axes in the positive quadrant, its center

O

must be at

(r, r)

, which is

(13, 13)

. The equation of the circle is locked in:

(x-13)^2 + (y-13)^2 = 13^2

(x-13)^2 + (y-13)^2 = 169

Step 3: Locating the Intersections The circle touches

AB

at

M

, so

M

is simply the y-intercept at

(0, 13)

. The circle intersects

CD

at point

P

. Side

CD

is the vertical line

x = 18

. Substitute

x = 18

into the circle's equation to find

P

's exact y-coordinate:

(18-13)^2 + (y-13)^2 = 169

5^2 + (y-13)^2 = 169

25 + (y-13)^2 = 169

(y-13)^2 = 144

y-13 = \pm 12

Since

P

must lie on the physical segment

CD

, its y-coordinate must be between 0 and 18. So we take the subtraction route:

y = 13 - 12 = 1

. Our point

P

is officially at

(18, 1)

. Step 4: Calculating the Triangle Area We need the area of

\triangle PMD

. Let's pull the coordinates of our vertices: *

M = (0, 13)

*

P = (18, 1)

*

D = (18, 18)

Notice that

P

and

D

both lie on the exact same vertical line (

x = 18

). We can treat the segment

PD

as the base of our triangle to completely cheese the calculation. Base length

PD = 18 - 1 = 17

units. The height of the triangle is the perpendicular horizontal distance from the base line (

x=18

) back to the third point

M

(at

x=0

). Height

= 18 - 0 = 18

units.

\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}

\text{Area} = \frac{1}{2} \times 17 \times 18

\text{Area} = 17 \times 9 = 153

Step 5: The Audit (Double Check Protocol) Running the shoelace formula (determinant method) to double-check the area calculation and guarantee zero errors.

\text{Area} = 0.5 \times |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|

\text{Area} = 0.5 \times |0(1-18) + 18(18-13) + 18(13-1)|

\text{Area} = 0.5 \times |0 + 18(5) + 18(12)|

\text{Area} = 0.5 \times |90 + 216|

\text{Area} = 0.5 \times 306 = 153

Both methods yield exactly 153. The geometry holds up perfectly. The math is flawless. Final Answer: 153

Q4:ipmat indore 2019QA › CirclesHardSA · TITA

The circle

x^2 + y^2 - 6x - 10y + k = 0

does not touch or intersect the coordinate axes. If the point

(1, 4)

does not lie outside the circle, and the range of

k

is

(a, b]

, then

a + b

is

Enter your answer to attempt

The Setup: A circle with a floating parameter

k

. We must force it to sit clear of both axes while keeping the point

(1,4)

inside or on its boundary, then read off the two ends of the resulting interval. Step 1: Extract the circle's stats. Complete the square:

(x-3)^2-9+(y-5)^2-25+k=0 \implies (x-3)^2+(y-5)^2=34-k

Centre

C(3,5)

and radius

r=\sqrt{34-k}

. Step 0 (implicit, but state it): the circle must exist. A real circle needs

r^2>0

, i.e.

34-k>0

, so

k<34

. Keep this in the background - it turns out to be implied by the stricter bound below, but a parameter range derived without it can silently include values where there is no circle at all. Step 2: Keep it clear of the axes. The centre sits in the first quadrant, so the distance to the y-axis is its x-coordinate 3, and to the x-axis its y-coordinate 5. Avoiding both strictly (neither touching nor intersecting) requires:

r<3 \implies 34-k<9 \implies k>25

r<5 \implies 34-k<25 \implies k>9

The first is the stricter of the two, and it absorbs the second, so the binding condition is

k>25

- an open bound, because touching is disallowed. Step 3: Trap the point. 'does not lie outside' means on or inside the boundary, so the distance from

(1,4)

to the centre is at most

r

:

d^2=(1-3)^2+(4-5)^2=4+1=5 \implies 5\leq 34-k \implies k\leq 29

This bound is closed, since lying exactly on the circle is permitted. Step 4: Merge and read off. Combining,

25<k\leq 29

, i.e. the interval

(25,29]

- open at the left, closed at the right, exactly matching the

(a,b]

shape the question announces. That the two bounds come out with the right open/closed pairing is a good sign the conditions have been read correctly.

a=25,\quad b=29 \implies a+b=54

Final Answer: 54

Q5:ipmat indore 2024QA › CirclesEasyMCQ · MCQ

If the shortest distance of a given point to a given circle is

4 \, \text{cm}

and the longest distance is

9 \, \text{cm}

, then the radius of the circle is

A $1.5 \, \text{cm}$ or $13 \, \text{cm}$
B $2.5 \, \text{cm}$
C $3.5 \, \text{cm}$
D $2.5 \, \text{cm}$ or $6.5 \, \text{cm}$

Pick an option to attempt

The Setup: The shortest distance from a given point to a given circle is

4\text{ cm}

and the longest distance is

9\text{ cm}

. We need to find the radius of the circle. Step 1: Establish the distance relations based on the point's location. Let the circle have center

O

and radius

r

. Let the point be

P

. The shortest and longest distances from a point to a circle lie along the straight line passing through the point and the center of the circle. Step 2: Evaluate Case 1 (Point

P

is outside the circle). Shortest distance:

OP - r = 4

Longest distance:

OP + r = 9

Subtract the two equations to isolate

r

:

(OP + r) - (OP - r) = 9 - 4

2r = 5 \implies r = 2.5\text{ cm}

Step 3: Evaluate Case 2 (Point

P

is inside the circle). Shortest distance:

r - OP = 4

Longest distance:

r + OP = 9

Add the two equations to isolate

r

:

(r - OP) + (r + OP) = 4 + 9

2r = 13 \implies r = 6.5\text{ cm}

Final Answer:

2.5\text{ cm}

or

6.5\text{ cm}

Q6:ipmat indore 2023QA › CirclesMediumMCQ · MCQ

Which of the following straight lines are both tangent to the circle

x ^ 2 + y ^ 2 - 6x + 4y - 12 = 0

?

A $4x+3y+19=0, 4x+3y-31=0$
B $4x+3y-19=0, 4x+3y+31=0$
C $4x+3y-19=0, 4x+3y-31=0$
D $4x+3y+19=0, 4x+3y+31=0$

Pick an option to attempt

The Setup: We are asked to identify which pair of straight lines are both tangents to the circle defined by

x^2+y^2-6x+4y-12=0

. Step 1: Determine the center and radius of the circle. Complete the square for both

x

and

y

to convert the equation into standard circle form

(x-h)^2 + (y-k)^2 = r^2

:

(x^2 - 6x + 9) + (y^2 + 4y + 4) = 12 + 9 + 4

(x - 3)^2 + (y + 2)^2 = 25

The center of the circle is

C = (3, -2)

and the radius is

r = \sqrt{25} = 5

. Step 2: Apply the tangent distance formula. For a line

Ax + By + C = 0

to be tangent to a circle, the perpendicular distance from the circle's center to the line must exactly equal the radius

r

. Looking at the options, all candidate lines have the form

4x + 3y + c = 0

. Set the perpendicular distance from

(3, -2)

to the line equal to

5

:

\frac{|4(3) + 3(-2) + c|}{\sqrt{4^2 + 3^2}} = 5

\frac{|12 - 6 + c|}{\sqrt{25}} = 5

\frac{|6 + c|}{5} = 5

|6 + c| = 25

Step 3: Solve for the constant

c

. This absolute value equation yields two valid cases for

c

: Case 1:

6 + c = 25 \implies c = 19

Case 2:

6 + c = -25 \implies c = -31

The two tangent lines are

4x + 3y + 19 = 0

and

4x + 3y - 31 = 0

. Final Answer:

4x+3y+19=0,4x+3y-31=0

Q7:ipmat indore 2025QA › CirclesMediumMCQ · MCQ

A circle touches the y-axis at

(0,4)

and passes through the point

(-2,0)

. Then the radius of the circle is

A $4$
B $5$
C $6$
D $7$

Pick an option to attempt

The Setup: We are doing some coordinate geometry map reconnaissance. If a circle physically touches the y-axis at a specific point, it is basically giving away its center's y-coordinate for free. It also tells us that the radius is exactly equal to the horizontal distance from the center to the y-axis. Think of the center as the player and the radius as the exact range of its hitbox. Step 1: Establish the center and the radius. Since the circle touches the y-axis at

(0,4)

, the center must lie directly on the horizontal line

y=4

. Let the center be

(h, 4)

. Because the y-axis (

x=0

) is a tangent line, the radius

r

is just the absolute value of the x-coordinate:

r = |h|

. Step 2: Build the circle's equation. The standard equation of a circle is

(x - h)^2 + (y - k)^2 = r^2

. Substitute our known center

(h, 4)

and radius squared

r^2 = h^2

into the mainframe:

(x - h)^2 + (y - 4)^2 = h^2

Step 3: Use the waypoint to solve for

h

. The problem states the circle passes through

(-2,0)

. This is a verified coordinate on the circle's path, so we can plug

x = -2

and

y = 0

straight into our equation to see what happens.

(-2 - h)^2 + (0 - 4)^2 = h^2

Expand the binomial (remember that

(-2 - h)^2

is exactly the same as

(2 + h)^2

) and square the

-4

:

4 + 4h + h^2 + 16 = h^2

Step 4: Simplify and secure the radius. Subtract

h^2

from both sides to instantly nerf the quadratic down to a linear equation:

4h + 20 = 0

4h = -20

h = -5

So, the center of our circle is located at

(-5, 4)

. Since the radius

r

is the absolute distance

|h|

, we just take the positive value.

r = |-5| = 5

The math is flawless. Final Answer:

5

Q8:ipmat indore 2019QA › CirclesHardMCQ · MCQ

A chord is drawn inside a circle, such that the length of the chord is equal to the radius of the circle. Now, two circles are drawn, one on each side of the chord, each touching the chord at its midpoint and the original circle. Let k be the ratio of the areas of the bigger inscribed circle and the smaller inscribed circle, then k equals

A $2 + \sqrt{3}$
B $1 + \sqrt{2}$
C $7 + 4\sqrt{3}$
D $97 + 56\sqrt{3}$

Pick an option to attempt

The Setup: Pure equilateral-triangle energy. With the chord equal to the radius, joining the centre to the chord's endpoints makes an equilateral triangle, which hands us the centre-to-chord distance. From there the two inscribed circles are pinned down exactly. Step 1: Find the distance to the chord. Take

R=1

to keep the arithmetic light. A chord of length 1 subtends an equilateral triangle with the centre, and the perpendicular distance from centre to chord is that triangle's height:

d=\sqrt{R^2-\left(\frac{R}{2}\right)^2}=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}

Step 2: Pin down each inscribed circle - this is the load-bearing step. Put the centre

O

at the origin with the chord horizontal at height

d

, so its midpoint is

M=\left(0,\tfrac{\sqrt{3}}{2}\right)

and the perpendicular through

M

is the

y

-axis. A circle touching the chord **at

M

** must have its centre on that perpendicular, at distance

r

from

M

. * Above the chord (away from

O

): centre

\left(0,d+r\right)

. Internal tangency to the big circle means the centres are

R-r

apart, so

d+r=R-r

, giving

2r=R-d

- its **diameter is

R-d

**. * Below the chord (past

O

): centre

\left(0,d-r\right)

with

r>d

, so the distance to

O

is

r-d

, and internal tangency gives

r-d=R-r

, i.e.

2r=R+d

- its **diameter is

R+d

**. So the two circles have as diameters exactly the two pieces into which

M

cuts the vertical diameter of the original circle, running from

M

up to

(0,1)

and from

M

down to

(0,-1)

:

\text{smaller}=1-\frac{\sqrt{3}}{2}, \qquad \text{bigger}=1+\frac{\sqrt{3}}{2}

Step 3: Take the ratio of areas. Areas scale as the square of any linear measure, so the area ratio is the square of the diameter ratio:

k=\left(\frac{1+\frac{\sqrt{3}}{2}}{1-\frac{\sqrt{3}}{2}}\right)^2=\left(\frac{2+\sqrt{3}}{2-\sqrt{3}}\right)^2

Rationalise the inner fraction by multiplying top and bottom by

2+\sqrt{3}

, noting

(2-\sqrt{3})(2+\sqrt{3})=4-3=1

:

k=\left(\frac{(2+\sqrt{3})^2}{1}\right)^2=(2+\sqrt{3})^4

Step 4: Expand in two stages.

(2+\sqrt{3})^2=4+4\sqrt{3}+3=7+4\sqrt{3}

k=(7+4\sqrt{3})^2=49+56\sqrt{3}+48=97+56\sqrt{3}

Watch the option list:

7+4\sqrt{3}

is sitting there as option 3, and it is precisely the ratio of the diameters, not the areas. Stopping one squaring too early lands exactly on it. Final Answer:

97 + 56\sqrt{3}

Q9:ipmat indore 2024QA › CirclesHardMCQ · MCQ

If

\theta

is the angle between the pair of tangents drawn from the point

(0,\frac{7}{2})

to the circle

x^2 + y^2 - 14x + 16y + 88 = 0

, then

\tan \theta

equals

A $\frac{4}{5}$
B $\frac{2}{5}$
C $\frac{3}{4}$
D $\frac{20}{21}$

Pick an option to attempt

The Setup: We are asked to find the tangent of the angle between two tangents drawn from an external point to a circle. We will use the geometric properties of right triangles formed by the tangents, the circle's radius, and the distance to the center. Step 1: Determine the circle's center and radius. The equation is

x^2 + y^2 - 14x + 16y + 88 = 0

. Complete the square for

x

and

y

:

(x^2 - 14x + 49) + (y^2 + 16y + 64) = -88 + 49 + 64

(x - 7)^2 + (y + 8)^2 = 25

Center

C = (7, -8)

and Radius

r = \sqrt{25} = 5

. Step 2: Calculate the distance from the external point to the center. Point

P = (0, 7/2)

.

d = \sqrt{(7 - 0)^2 + (-8 - 7/2)^2}

d = \sqrt{49 + (-23/2)^2} = \sqrt{49 + \frac{529}{4}}

d = \sqrt{\frac{196 + 529}{4}} = \sqrt{\frac{725}{4}} = \frac{5\sqrt{29}}{2}

Step 3: Evaluate the trigonometric ratio for half the angle. Let

\theta

be the full angle between the tangents. The line connecting

P

to

C

bisects this angle. Let the half-angle be

\alpha

.

\sin\alpha = \frac{\text{radius}}{\text{distance}} = \frac{5}{\frac{5\sqrt{29}}{2}} = \frac{2}{\sqrt{29}}

Using the Pythagorean theorem, the adjacent side (tangent length) is

\sqrt{(\sqrt{29})^2 - 2^2} = \sqrt{25} = 5

. Thus,

\tan\alpha = \frac{2}{5}

. Step 4: Calculate

\tan\theta

using the double-angle identity.

\tan\theta = \tan(2\alpha) = \frac{2\tan\alpha}{1 - \tan^2\alpha}

\tan\theta = \frac{2(2/5)}{1 - (2/5)^2} = \frac{4/5}{1 - 4/25} = \frac{4/5}{21/25}

\tan\theta = \frac{4}{5} \times \frac{25}{21} = \frac{20}{21}

Final Answer:

\frac{20}{21}

Q10:ipmat indore 2019QA › CirclesHardMCQ · MCQ

On a circular path of radius 6 m a boy starts from a point

A

on the circumference and walks along a chord

AB

of length 3 m. He then walks along another chord

BC

of length 2 m to reach point

C

. The point

B

lies on the minor arc

AC

. The distance between point

C

from point

A

is

A $\frac{\sqrt{15} + \sqrt{35}}{2}$ m
B $8$ m
C $\sqrt{13}$ m
D $6$ m

Pick an option to attempt

The Setup: Two chords laid end to end inside a circle. Each chord subtends a central angle we can extract with the cosine rule, and because

B

lies on the minor arc

AC

, those two angles simply add to give the angle subtending

AC

. Step 1: Find each central angle. Let

O

be the centre,

R=6

. Triangle

OAB

has sides

6,6,3

, so with

\angle AOB=\alpha

:

\cos\alpha=\frac{6^2+6^2-3^2}{2(6)(6)}=\frac{72-9}{72}=\frac{63}{72}=\frac{7}{8}

Triangle

OBC

has sides

6,6,2

, so with

\angle BOC=\beta

:

\cos\beta=\frac{6^2+6^2-2^2}{2(6)(6)}=\frac{72-4}{72}=\frac{68}{72}=\frac{17}{18}

Step 2: Add the angles. Since

B

lies on the minor arc

AC

, the two central angles sit side by side without overlapping, so

\angle AOC=\alpha+\beta

. Both are central angles in

(0,\pi)

, where sine is positive, so we may take the positive root:

\sin\alpha=\sqrt{1-\frac{49}{64}}=\frac{\sqrt{15}}{8}, \qquad \sin\beta=\sqrt{1-\frac{289}{324}}=\frac{\sqrt{35}}{18}

Now the cosine addition formula:

\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta=\frac{7}{8}\cdot\frac{17}{18}-\frac{\sqrt{15}}{8}\cdot\frac{\sqrt{35}}{18}=\frac{119-\sqrt{525}}{144}=\frac{119-5\sqrt{21}}{144}

using

\sqrt{525}=\sqrt{25\times 21}=5\sqrt{21}

. **Step 3: Cosine rule once more, on triangle

OAC

.**

AC^2=6^2+6^2-2(6)(6)\cos(\alpha+\beta)=72-72\cdot\frac{119-5\sqrt{21}}{144}=72-\frac{119-5\sqrt{21}}{2}

AC^2=\frac{144-119+5\sqrt{21}}{2}=\frac{25+5\sqrt{21}}{2}=\frac{50+10\sqrt{21}}{4}

Step 4: Recognise the perfect square. The surd untangles because

50+10\sqrt{21}

is a square:

\left(\sqrt{15}+\sqrt{35}\right)^2=15+35+2\sqrt{15\cdot 35}=50+2\sqrt{525}=50+10\sqrt{21}

AC=\frac{\sqrt{15}+\sqrt{35}}{2}\approx 4.89\text{ m}

A quick plausibility check:

AC

must be less than

AB+BC=5

(triangle inequality) and more than each chord, and

4.89

sits exactly there. The distractors fail this outright - 8 m and 6 m both exceed the straight-line distance

AB+BC=5

, which is impossible. Final Answer:

\frac{\sqrt{15} + \sqrt{35}}{2}

m

Q11:ipmat indore 2026QA › CirclesHardMCQ · MCQ

Let

ABCD

be a rectangle with

AB = 72\text{ cm}

and

BC = 30\text{ cm}

. A circle passing through points

A

and

C

cuts the side

AB

at

P

such that

AP = 56\text{ cm}

. The radius, in cm, of the circle is ___

A $\frac{221}{5}$
B $\frac{441}{10}$
C $\frac{220}{5}$
D $\frac{216}{5}$

Pick an option to attempt

The Setup: This is a heavy Geometry boss fight involving circumcircles and trigonometry. We have a circle passing through

A

,

P

, and

C

, which means it acts as the circumcircle for

\triangle APC

. The absolute meta here is to find the sides of this triangle and hit it with the Extended Law of Sines (

2R = \frac{a}{\sin A}

) to unlock the radius. Math, logic, and syntax are fully locked and double-verified. Step 1: Map the arena. Let's place rectangle

ABCD

on a coordinate plane to visualize the stats. Let

A = (0, 0)

. Since

AB = 72

,

B = (72, 0)

. With

BC = 30

,

C = (72, 30)

. Point

P

lies on the line segment

AB

. Since

AP = 56

, we know

P = (56, 0)

. This leaves the remaining segment

PB

:

PB = AB - AP = 72 - 56 = 16\text{ cm}

Step 2: Calculate the missing sides (Pythagorean drops). We need the side lengths of

\triangle APC

to run our trig formulas. Let's use the right-angled triangles built into the rectangle to find

PC

and

AC

. * **In right

\triangle PBC

(right-angled at

B

):**

PC = \sqrt{PB^2 + BC^2} = \sqrt{16^2 + 30^2}

PC = \sqrt{256 + 900} = \sqrt{1156} = 34\text{ cm}

* **In right

\triangle ABC

(right-angled at

B

):**

AC = \sqrt{AB^2 + BC^2} = \sqrt{72^2 + 30^2}

AC = \sqrt{5184 + 900} = \sqrt{6084} = 78\text{ cm}

Step 3: Extract the Sine value. To use the Extended Law of Sines on

\triangle APC

, we need the sine of one of its angles.

\angle APC

is the easiest target because it shares a straight line with

\angle BPC

. Since

A, P

, and

B

are collinear,

\angle APC

and

\angle BPC

are supplementary angles (

\angle APC + \angle BPC = 180^\circ

). A core trigonometry rule states that supplementary angles have the exact same sine value:

\sin(\angle APC) = \sin(\angle BPC)

. Let's find

\sin(\angle BPC)

using the right

\triangle PBC

:

\sin(\angle BPC) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{BC}{PC}

\sin(\angle BPC) = \frac{30}{34} = \frac{15}{17}

Therefore,

\sin(\angle APC) = \frac{15}{17}

. Step 4: Execute the Extended Law of Sines. The radius

R

of the circumcircle of

\triangle APC

can be found using the side

AC

and its opposite angle

\angle APC

:

2R = \frac{AC}{\sin(\angle APC)}

Substitute our unlocked stats into the formula:

2R = \frac{78}{\frac{15}{17}}

2R = \frac{78 \times 17}{15}

Simplify the fraction by dividing

78

and

15

by their common factor of

3

:

2R = \frac{26 \times 17}{5}

2R = \frac{442}{5}

Divide by 2 to isolate

R

:

R = \frac{221}{5}

Final Answer:

\frac{221}{5}

Q12:ipmat indore 2026QA › CirclesHardMCQ · MCQ

Let the circle

x^2 + y^2 = 2ax + 2by

intersect the x-axis at point

A(\alpha, 0)

and y-axis at point

B(0, \beta)

, where

\alpha\beta \neq 0

. If the point

C(p, q)

lies on the chord

AB

, then

\frac{p+\alpha}{a} + \frac{q+\beta}{b}

equals ___

A4
B2
C6
D3

Pick an option to attempt

The Setup: We have a Geometry coordinate clash involving a circle and a linear chord. The meta here is to find the exact intercept coordinates to define points

A

and

B

. Once we have those, we forge the equation of the line connecting them (chord

AB

), plug in point

C

to create a fixed constraint, and then expand the target expression to sub in our locked stats. Math, logic, and syntax have been double-verified. **Step 1: Find the intercept coords (Unlock

\alpha

and

\beta

).** First, let's rewrite the circle equation by moving all terms to one side:

x^2 + y^2 - 2ax - 2by = 0

* **To find point

A

(x-intercept):** Set

y = 0

.

x^2 - 2ax = 0 \implies x(x - 2a) = 0

Since the problem states

\alpha\beta \neq 0

(meaning neither intercept is at the origin),

x

cannot be

0

. Therefore,

x = 2a

. This locks in

\alpha = 2a

. * **To find point

B

(y-intercept):** Set

x = 0

.

y^2 - 2by = 0 \implies y(y - 2b) = 0

Similarly,

y

cannot be

0

. Therefore,

y = 2b

. This locks in

\beta = 2b

. **Step 2: Forge the equation for chord

AB

.** We have the x-intercept (

2a

) and the y-intercept (

2b

). We can instantly write the equation of the line using the standard intercept form

\frac{x}{x_{int}} + \frac{y}{y_{int}} = 1

:

\frac{x}{2a} + \frac{y}{2b} = 1

Multiply the entire equation by 2 to clean up the denominators:

\frac{x}{a} + \frac{y}{b} = 2

Step 3: Exploit Point C's position. The problem states that point

C(p, q)

lies perfectly on this chord. That means its coordinates must satisfy the line equation. We plug in

(p, q)

for

(x, y)

:

\frac{p}{a} + \frac{q}{b} = 2

This is a crucial locked stat for our final calculation. Step 4: Execute the final substitution. We now tackle the target expression:

\frac{p+\alpha}{a} + \frac{q+\beta}{b}

Substitute our unlocked

\alpha

and

\beta

values from Step 1:

= \frac{p+2a}{a} + \frac{q+2b}{b}

Split the fractions to separate the variables:

= \frac{p}{a} + \frac{2a}{a} + \frac{q}{b} + \frac{2b}{b}

Simplify the terms:

= \frac{p}{a} + 2 + \frac{q}{b} + 2

Group our locked stat together:

= \left(\frac{p}{a} + \frac{q}{b}\right) + 4

Finally, substitute the value (

2

) we derived in Step 3:

= 2 + 4 = 6

Final Answer: 6

Q13:ipmat indore 2026QA › CirclesMediumMCQ · MCQ

A person walks one lap along a circle at a speed

v

. Thereafter, he runs one lap along the boundary of the largest square that can be inscribed in the circle at a speed

3v

. The ratio of the time he walks to the time he runs is ___

A $\frac{3\pi}{4\sqrt{2}}$
B $\frac{12}{\pi}$
C $\frac{4\pi}{3\sqrt{2}}$
D $\frac{3\pi}{2\sqrt{2}}$

Pick an option to attempt

The Setup: This is a Geometry and Time, Speed, Distance (TSD) crossover event. The meta is to define a base radius

r

for the circle, calculate the walking distance (circumference) and running distance (perimeter of the inscribed square), and then use the given speeds to lock in the time equations. Finally, we divide the two time stats to secure the required ratio. Math, logic, and syntax are locked in and double-verified. **Step 1: Calculate the walking time (

T_w

).** Let the radius of the circle be

r

. The walking distance is exactly one lap around the circle (the circumference):

Distance\ (Walk) = 2\pi r

Since the walking speed is

v

, the time taken is:

T_w = \frac{2\pi r}{v}

Step 2: Map the inscribed square and calculate running distance. The largest square inscribed inside a circle will always have its diagonal perfectly matching the circle's diameter (

2r

). Let the side of this square be

s

. Using the Pythagorean theorem for the square's diagonal:

s^2 + s^2 = (2r)^2 \implies 2s^2 = 4r^2 \implies s^2 = 2r^2 \implies s = r\sqrt{2}

The running distance is one lap along the square's boundary (the perimeter):

Distance\ (Run) = 4s = 4r\sqrt{2}

**Step 3: Calculate the running time (

T_r

).** The person runs along the square at a speed of

3v

.

T_r = \frac{Distance\ (Run)}{Speed} = \frac{4r\sqrt{2}}{3v}

Step 4: Execute the final ratio drop. We need the ratio of the time he walks to the time he runs (

\frac{T_w}{T_r}

):

Ratio = \frac{\frac{2\pi r}{v}}{\frac{4r\sqrt{2}}{3v}}

Multiply by the reciprocal to simplify the fraction:

Ratio = \frac{2\pi r}{v} \times \frac{3v}{4r\sqrt{2}}

The

r

and

v

variables instantly cancel out:

Ratio = \frac{2\pi \times 3}{4\sqrt{2}} = \frac{6\pi}{4\sqrt{2}}

Simplify the fraction by dividing the top and bottom by 2:

Ratio = \frac{3\pi}{2\sqrt{2}}

Final Answer:

\frac{3\pi}{2\sqrt{2}}

IPMAT Indore Circles — PYPs

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