The coordinates of the point from where the t | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the point be $(h, k)$. The length of the tangent from $(h, k)$ to a circle $S = 0$ is $\sqrt{S(h, k)}$.Since the lengths of the tangents are e

Vedclass · Accepted Answer

$\left( -2, -\frac{5}{2} \right)$

Vedclass · Answer

(B) Let the point be $(h, k)$. The length of the tangent from $(h, k)$ to a circle $S = 0$ is $\sqrt{S(h, k)}$.Since the lengths of the tangents are equal,the power of the point with respect to the three circles must be equal.Let $S_1 = x^2 + y^2 - 1 = 0$,$S_2 = x^2 + y^2 + 8x + 15 = 0$,and $S_3 = x^2 + y^2 + 10y + 24 = 0$.Equating $S_1 = S_2$:$x^2 + y^2 - 1 = x^2 + y^2 + 8x + 15$$-1 = 8x + 15$ $\Rightarrow 8x = -16$ $\Rightarrow x = -2$.Equating $S_1 = S_3$:$x^2 + y^2 - 1 = x^2 + y^2 + 10y + 24$$-1 = 10y + 24$ $\Rightarrow 10y = -25$ $\Rightarrow y = -\frac{5}{2}$.Thus,the required point is $\left( -2, -\frac{5}{2} \right)$.

The coordinates of the point from where the tangents drawn to the circles ${x^2} + {y^2} = 1$,${x^2} + {y^2} + 8x + 15 = 0$,and ${x^2} + {y^2} + 10y + 24 = 0$ are of the same length are:

Similar Questions

The radical axis of circles $x^2+y^2+5x+4y-5=0$ and $x^2+y^2-3x+5y-6=0$ is

If the radical axis of the circles $x^2+y^2+2gx+2fy+c=0$ and $2x^2+2y^2+3x+8y+2c=0$ touches the circle $x^2+y^2+2x+2y+1=0$,then

Two circles of equal radius $a$ cut orthogonally. If their centres are $(2, 3)$ and $(5, 6)$,then the radical axis of these circles passes through the point:

If $d$ is the distance between the centres of two circles,$r_1$ and $r_2$ are their radii,and $d = r_1 + r_2$,then

The equation of the circle which passes through the intersection of ${x^2} + {y^2} + 13x - 3y = 0$ and $2{x^2} + 2{y^2} + 4x - 7y - 25 = 0$ and whose centre lies on $13x + 30y = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module