lf the tangents from $P (h, k)$ 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

2 h + 10 k + 11 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 h - 10 k + 11 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

10 h - 2 k + 11 = 0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

10 h + 2 k + 11 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D $10 h - 2 k + 11 = 0$
$s_{1} : x^{2} + y^{2} - 4 x - 5 = 0$ and $s_{2} : x^{2} + y^{2} + 6 x - 2 y + 6 = 0$
radical axis of given circle is $s_{1} - s_{2} = 0$
$- 10 x + 2 y - 11 = 0$
$10 x - 2 y + 11 = 0$
So, radical axis is a locus of point from which we can
draw tangent of equal length to circles.
So, $P (h, k)$ lie on radical axis of $s_{1}$ and $s_{2}$
$\Rightarrow 10 h - 2 k + 11 = 0$

Suggest Corrections

Similar questions

Q. 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

10 a - 2 b

is equal to

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

Q. The point, the length of the tangents from which to the following three circles

x^{2} + y^{2} + 4 x + 6 y - 19 = 0

x^{2} + y^{2} - 2 x - 2 y - 5 = 0

and

x^{2} + y^{2} - 9 = 0

are equal is

Q. If the circle

x^{2} + y^{2} + 8 x - 4 y + c = 0

touches the circle

x^{2} + y^{2} + 2 x + 4 y - 11 = 0

externally and cuts the circle

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

orthogonally then

k =

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

Join BYJU'S Learning Program

lf the tangents from P(h,k) to the circles x2+y2−4x−5=0 andx2+y2+6x−2y+6=0 are equal then

lf the tangents from $P (h, k)$ 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