JEE Advanced 2020 Paper 1 · Mathematics · Straight Lines and Circles · medium

The circles C1: x^2+y^2-2x-4y-4=0 and C2: x^2+y^2-8x-12y+36=0 intersect at two points A and B. A third circle is drawn with the common chord AB as a…

Numerical answer

The circles

C_1:\ x^{2}+y^{2}-2x-4y-4=0

and

C_2:\ x^{2}+y^{2}-8x-12y+36=0

intersect at two points

A

and

B

. A third circle is drawn with the common chord

AB

as a diameter. Find the area of this circle. Give the answer correct to two decimal places.

Show answer and solution

Answer: 18.10

C_1

has centre

O_1(1,2)

and radius

r_1=\sqrt{1+4+4}=3

C_2

has centre

O_2(4,6)

and radius

r_2=\sqrt{16+36-36}=4

. The distance

O_1O_2=\sqrt{9+16}=5

and

r_1^{2}+r_2^{2}=25=O_1O_2^{2}

, so the circles meet orthogonally and

\triangle O_1AO_2

is right-angled at

A

. The common chord

AB

is perpendicular to

O_1O_2

, and half of it is the altitude from

A

to the hypotenuse:

\frac{AB}{2}=\frac{r_1r_2}{O_1O_2}=\frac{12}{5}=2.4

. So

AB=4.8

and the new circle has radius

2.4

. Its area is

\pi(2.4)^{2}=5.76\pi\approx18.0956

, i.e.

18.10

. The radical axis

6x+8y-40=0

gives the same chord length by direct computation.

Original question written in the JEE Advanced 2020 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake