The equation of the image of the circle $x^{2} + y^{2} - 6 x - 4 y + 12 = 0$ by the line mirror $x + y - 1 = 0$ is

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

$x^{2} + y^{2} + 2 x + 4 y - 4 = 0$

$x^{2} + y^{2} + 2 x + 4 y + 4 = 0$

$x^{2} + y^{2} - 2 x + 4 y + 4 = 0$

$x^{2} + y^{2} + 2 x - 4 y + 4 = 0$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

The given circle is $x^{2} + y^{2} - 6 x - 4 y + 12 = 0$

Center and radius of the above circle is $C (3, 2)$ and $r = \sqrt{9 + 4 - 12} = 1$

To get the image of the circel in the line mirror $x + y - 1 = 0$ , find the image of the point $C (3, 2)$ with respect to the line $x + y - 1 = 0$

Let $(h, k)$ be the image of $(3, 2)$ with respect to the line $x + y - 1 = 0$

$\frac{h - 3}{1} = \frac{k - 2}{1} = - 2 (\frac{3 + 2 - 1}{1 + 1})$

Hence, $\frac{h - 3}{1} = - 4 \Rightarrow h - 3 = - 4 \Rightarrow h = - 1$

and $\frac{k - 2}{1} = - 4 \Rightarrow k = - 2$

New center is (-1,-2)

The equation of image circle is ${(x + 1)}^{2} + {(y + 2)}^{2} = 1 \Rightarrow x^{2} + y^{2} + 2 x + 4 y + 4 = 0$

Watch 3-min video & get full concept clarity