Question

The locus of the centre of all circles passing through (2, 1) and cutting the circle $x^2+y^2=9$ orthogonally is

• A. x + y = 3
• B. 2x + y = 7
• C. x - y = 1
• D. x + 2y = 3

Answer 1

Answer: (B)

2x + y = 7

Let the centre of circle be $(h,k)$ then eqn. of circles can be written as,

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

$\Rightarrow x^2+y^2-2hx-2ky+h^2+k^2=r^2$

$\Rightarrow x^2+y^2-2hx-2ky+h^2+k^2-r^2=0$ ......... $\left(1\right)$

Since the circle passes through $(2,1)$, so the circles should satisfy the eqn.

$4+1-4h-2k+h^2+k^2-r^2=0$

$\Rightarrow 5-4h-2k+h^2+k^2-r^2=0$ ....... $\left(2\right)$

Now, circle $\left(1\right)$ and $x^2+y^2=9$ cuts orthogonally.

Hence, by condition of orthogonality

$2a_1{a}_2+2f_1{f}_2=c_1+c_2$

Since, $a_2=0$ and $f_2=0$

$c_1+c_2=0$

$h^2+k^2-r^2=9$ $⟶\left(3\right)$

Putting (3) in eqn. (2)

$5-4h-2k+9=0$

$\Rightarrow 4h+2k=14$

$\Rightarrow 2h+k=7$

Now, placing $h=x,$ & $k=y$

$2x+y=7$ Reqd. eqn.