Question

A circle passes through the origin and has its centre on $y = x.$ If it cuts $x^{2} + y^{2} - 4x - 6y + 10 = 0$ orthogonally, then the equation of the circle is

• A. $x^{2} + y^{2} - x - y = 0$
• B. $x^{2} + y^{2} - 6x - 4y = 0$
• C. $x^{2} + y^{2} - 2x - 2y = 0$
• D. $x^{2} + y^{2} + 2x + 2y = 0$

Answer 1

Answer: (C)

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

Answer 2

Let the required circle be $x^{2} + y^{2} + 2gx + 2fy + c = 0$.......(i)

This passes through (0, 0), therefore c = 0

The centre $( - g, - f)$ of (i) lies on y = x, hence g = f.

Since (i) cuts the circle $x^{2} + y^{2} - 4x - 6y + 10 = 0$ orthogonally, therefore $2( - 2g - 3f) = c + 10$

$\Rightarrow - 10g = 10$

$\Rightarrow g = f = - 1$

$(\because g = f$ and $c = 0)$. Hence the required circle is

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