The lengths of the tangents from any point on the circle $x^2 + y^2 + 8x + 1 = 0$ to the circles $x^2 + y^2 + 7x + 1 = 0$ and $x^2 + y^2 + 4x + 1 = 0$ are in the ratio
Step-by-Step Solution
Key Concept: The length of tangent from an external point to a circle is $\sqrt{S}$ where $S$ is obtained by substituting the point coordinates into the circle equation.
Let the point on $x^2 + y^2 + 8z + 1 = 0$ be $(h, k)$. Then $h^2 + k^2 + 8h + 1 = 0$. The ratio of lengths of tangents from $(h, k)$ to the circles $x^2 + y^2 + 7x + 1 = 0$ and $x^2 + y^2 + 4x + 1 = 0$ is found using the tangent length formula $\sqrt{S}$ where $S$ is the value obtained by substituting the point into the circle equation. Computing $\sqrt{\frac{h^2+k^2+7h+1}{h^2+k^2+4h+1}} = \sqrt{\frac{-h+7h}{-4h+7h}} = \frac{1}{2} = 1:2$.
Correct Answer: 1