JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Circle

Let the tangent to the circle

x^{2} + y^{2} = 25

at the point

R (3, 4)

meet

x

-axis and

y

-axis at points

P

and

Q

, respectively. If

r

is the radius of the circle passing through the origin

O

and having centre at the incentre of the triangle OPQ , then

r^{2}

is equal to:

Answer: (D)

Solution

The tangent to

x^{2} + y^{2} = 25

R (3, 4)

3 x + 4 y = 25

, meeting the axes at

P (\frac{25}{3}, 0)

and

Q (0, \frac{25}{4})

. Triangle

O P Q

is right-angled at the origin, with legs

O P = \frac{25}{3}

and

O Q = \frac{25}{4}

and hypotenuse

P Q = 25 \frac{1}{9} + \frac{1}{16} = 25 \cdot \frac{5}{12} = \frac{125}{12}

For a right triangle the inradius is

ρ = \frac{O P + O Q - P Q}{2} = \frac{1}{2} (\frac{25}{3} + \frac{25}{4} - \frac{125}{12}) = \frac{25}{12}

.
Both legs lie along the axes and the triangle sits in the first quadrant, so the incentre is

(\frac{25}{12}, \frac{25}{12})

. The circle centred there and passing through the origin has

r^{2} = (\frac{25}{12})^{2} + (\frac{25}{12})^{2} = \frac{625}{72}

r^{2} = \frac{625}{72}

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Single correct option