If a circle S passing through the point (3,4) cuts the circle x2+y2=36 orthogonally, then the locus of the centre of S is

Question

If a circle S passing through the point (3,4) cuts the circle x2+y2=36 orthogonally, then the locus of the centre of S is (A) x2+y2-6 x-8 y+11=0 (B) 6 x+8 y-61=0 (C) x2+y2-8 x-6 y+11=0 (D) 6 x+8 y+11=0

Tardigrade Admin · Accepted Answer

Let the circle is x2+y2+2 g x+2 f y +c=0, having centre (-g,-f), since it passes through the point (3,4) So, 9+16+6 g+8 f + c=0 ...(i) And circle is intersecting the other circle So x2+y2=36 orthogonally, so 2 g(0)+2 f(0) =c-36 ⇒ C =36 ...(ii) From Eqs. (i) and (ii) -6 g-8 f=61 Now, on taking locus of point (-g,-f), we are getting 6 x+8 y-61=0

Q. If a circle $S$ passing through the point $(3, 4)$ cuts the circle $x^{2} + y^{2} = 36$ orthogonally, then the locus of the centre of $S$ is

$x^{2} + y^{2} - 6 x - 8 y + 11 = 0$

$6 x + 8 y - 61 = 0$

$x^{2} + y^{2} - 8 x - 6 y + 11 = 0$

$6 x + 8 y + 11 = 0$

Q. If a circle S passing through the point (3,4) cuts the circle x2+y2=36 orthogonally, then the locus of the centre of S is

x2+y2−6x−8y+11=0

6x+8y−61=0

x2+y2−8x−6y+11=0

6x+8y+11=0

Q. If a circle $S$ passing through the point $(3, 4)$ cuts the circle $x^{2} + y^{2} = 36$ orthogonally, then the locus of the centre of $S$ is

$x^{2} + y^{2} - 6 x - 8 y + 11 = 0$

$6 x + 8 y - 61 = 0$

$x^{2} + y^{2} - 8 x - 6 y + 11 = 0$

$6 x + 8 y + 11 = 0$