The tangent at A(-1, 2) on the circle x^2+y^2 | vedclass

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(B) Let circle $C_1: x^2+y^2-4x-8y+7=0$. The centre is $O_1(2, 4)$ and radius $r_1 = \sqrt{2^2+4^2-7} = \sqrt{13}$.Let circle $C_2: x^2+y^2+4x+6y=0$.

Vedclass · Accepted Answer

$\left(-\frac{1}{3}, 1\right)$

Vedclass · Answer

(B) Let circle $C_1: x^2+y^2-4x-8y+7=0$. The centre is $O_1(2, 4)$ and radius $r_1 = \sqrt{2^2+4^2-7} = \sqrt{13}$.Let circle $C_2: x^2+y^2+4x+6y=0$. The centre is $O_2(-2, -3)$ and radius $r_2 = \sqrt{(-2)^2+(-3)^2-0} = \sqrt{13}$.Since the radii are equal and the tangent at $A$ on $C_1$ touches $C_2$ at $B$,$B$ is the point of contact of the two circles. Since $r_1 = r_2$,$B$ is the midpoint of the line segment joining the centres $O_1$ and $O_2$.$B = \left(\frac{2-2}{2}, \frac{4-3}{2}\right) = \left(0, \frac{1}{2}\right)$.We need to find a point of trisection of the segment $AB$ where $A(-1, 2)$ and $B(0, 1/2)$.Using the section formula for a point dividing $AB$ in ratio $1:2$:$x = \frac{1(0) + 2(-1)}{1+2} = -\frac{2}{3}$,$y = \frac{1(1/2) + 2(2)}{1+2} = \frac{0.5+4}{3} = \frac{4.5}{3} = 1.5$.Using the section formula for a point dividing $AB$ in ratio $2:1$:$x = \frac{2(0) + 1(-1)}{2+1} = -\frac{1}{3}$,$y = \frac{2(1/2) + 1(2)}{2+1} = \frac{1+2}{3} = 1$.Thus,the point $\left(-\frac{1}{3}, 1\right)$ is a point of trisection.

The tangent at $A(-1, 2)$ on the circle $x^2+y^2-4x-8y+7=0$ touches the circle $x^2+y^2+4x+6y=0$ at $B$. Then,a point of trisection of $AB$ is

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