Tangents are drawn from the point P(1, 8) to the circle $x^{2} + y^{2} - 6 x - 4 y - 11 = 0$ touch the circle at the point A and B, then equation of the circumcircle of the triangle PAB is

If the ratio of the lengths of tangents from a point to the circles $x^{2} + y^{2} + 4 x + 3$ = 0, $x^{2} + y^{2} - 6 x + 5$ = 0 is 1:2 then the locus of P is a circle whose centre is

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If the length of the tangents from (a,b) to the circles x2+y2−4x−5=0 and x2+y2+6x−2y+6=0 are equal, then the value of 10a−2b is

The correct option is B −11Given circles are x2+y2−4x−5=0 and x2+y2+6x−2y+6=0 Point from where length of tangents are equal lies on the radical axis, So the radical axis is S1−S2=0⇒−10x+2y−11=0 Putting (a,b) in the equation of radical axis, we get ∴10a−2b=−11

If the length of the tangents from $(a, b)$ to the circles $x^{2} + y^{2} - 4 x - 5 = 0$ and $x^{2} + y^{2} + 6 x - 2 y + 6 = 0$ are equal, then the value of $10 a - 2 b$ is