Question

Centre of a circle S passing through the intersection points of circles x² + y² -6x =0 & x² + y² - 4y =0 lies on the line 2x -— 3y + 12 = 0 then circle S passes through

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

The given equations of the circles are :

${x}^{2} + {y}^{2} - 6x = 0 \: \: \: \: .......(1)$

${x}^{2} + {y}^{2} - 4y = 0 \: \: \: \: ......(2)$

Now the equation of the circle passing through the point of intersection of the above two circles is :

$( {x}^{2} + {y}^{2} - 6x) + k( {x}^{2} + {y}^{2} - 4y) = 0$

$\implies \: (1 + k) {x}^{2} + (1 + k) {y}^{2} - 6x - 4ky = 0$

$\implies \displaystyle \: \: {x}^{2} + {y}^{2} - \frac{6}{1 + k} x - \frac{4k}{1 + k} y = 0$

So the centre of the circle is

$\displaystyle \: \:( \frac{3}{1 + k} , \frac{2k}{1 + k} )$

It lies on the line

$2x - 3y + 12 = 0$

So

$\displaystyle \: \:2 \times \frac{3}{1 + k} - 3 \times \frac{2k}{1 + k} + 12 = 0$

$\implies \: 6 - 6k + 12 + 12k = 0$

$\implies \: 6k = - 18$

$\implies \: k = - 3$

So the equation of the circle is

$- 2 {x}^{2} - 2 {y}^{2} - 6x + 12y = 0$

$\implies \: {x}^{2} + {y}^{2} + 3x - 6y = 0$

Now ( - 3, 6) satisfies the above equation

So The above circle passes through the point $( - 3, 6)$

$\red{ \boxed{SO \: THE \: REQUIRED \: ANSWER \: IS \: ( - 3, 6) }}$