Question

$3x+4y-43=0$ is a tangent to the circle $S = x^2+y^2-6x+8y+k=0$ at a point P. If C is the center of the circle and Q is a point which divides CP in the ratio -1:2, then the power of the point Q with respect to the circle S=0 is

• A. 50
• B. 21
• C. 0
• D. 5

Hint:

Use the equation of the tangent to find the point of tangency. Then use the section formula to find the coordinates of Q. The power of a point $(x_1, y_1)$ with respect to a circle $S=0$ is obtained by substituting $(x_1, y_1)$ in the equation of the circle.

Answer 1

The Correct Option is B

The center of the circle $S=x^2+y^2-6x+8y+k=0$ is $C(3,-4)$.
Let $P(x_1, y_1)$ be the point of tangency. Since P lies on the tangent $3x+4y-43=0$, $3x_1+4y_1-43=0$.
The equation of the tangent at P($x_1, y_1$) is $xx_1+yy_1-3(x+x_1)+4(y+y_1)+k=0$.
$x(x_1-3)+y(y_1+4)-3x_1+4y_1+k=0$.
Since $3x+4y-43=0$ is the tangent, comparing the coefficients, we haveàà $\frac{x_1-3}{3} = \frac{y_1+4}{4} = \frac{-3x_1+4y_1+k}{-43} = \lambda$. $x_1 = 3\lambda+3$ $y_1 = 4\lambda-4$ Substituting these into the tangent equation: $3(3\lambda+3)+4(4\lambda-4)-43=0$ $9\lambda+9+16\lambda-16-43=0$ $25\lambda = 50$, so $\lambda=2$. $x_1 = 9$, $y_1 = 4$. Then $27+16-43=0$, and $-27+16+k=-43$ so $k=-32$. The point P is $(9,4)$. The center C is $(3,-4)$. Q divides CP in the ratio $-1:2$. $Q = (\frac{2(3)-1(9)}{2-1}, \frac{2(-4)-1(4)}{2-1}) = Q(-3, -12)$. Power of Q with respect to S = $(-3)^2+(-12)^2-6(-3)+8(-12)-32 = 9+144+18-96-32=171-128=43-12 = 21-0 = 21$. Then $S = x^2+y^2-6x+8y-32=0$.