Question

The equation of the circle passing through the point (–2, 4) and through the points of intersection of the circle $x^2+y^2-2x-6y+6=0$ and the line 3x + 2y - 5 = 0, is

• A. $x^2+y^2+2x-4y-4=0$
• B. $x^2+y^2+4x-2y-4=0$
• C. $x^2+y^2-3x-4y=0$
• D. $x^2+y^2-4x-2y-4=0$

Answer 1

The correct option is B $x^2+y^2+4x-2y-4=0$

Required equation of the circle,
$(x^2+y^2-2x-6y+6)+\lambda (3x+2y-5)=0,$

This circle passing through points (-2, 4), therefore (4 + 16 + 4 - 24 + 6) + $\lambda$ (-6 + 8 - 5) = 0, $\therefore \lambda$ = 2

$\therefore (x^2+y^2-2x-6y+6)+2(3x+2y-5)=0$
$\Rightarrow x^2+y^2+4x-2y-4=0.$