Question

If $\alpha$, $\beta$ are the angles between the two tangents drawn from (0, 0) and (8, 6) respectively to the circle $x^2 + y^2 - 14x + 2y + 25 = 0$, then $\alpha - \beta$ =

• A. $\frac{\pi}{2}$
• B. $\frac{\pi}{3}$
• C. 0°
• D. $\frac{\pi}{4}$

Answer 1

Answer: (C)

0°

Answer 2

The given circle is $x^2 + y^2 - 14x + 2y + 25 = 0$.

To find the center and radius of the circle, we compare it with the general equation of a circle $x^2 + y^2 + 2gx + 2fy + c = 0$.

Here, $2g = -14 \Rightarrow g = -7$.

$2f = 2 \Rightarrow f = 1$.

$c = 25$.

The center of the circle $C$ is $(-g, -f) = (7, -1)$.

The radius of the circle $r$ is $\sqrt{g^2 + f^2 - c} = \sqrt{(-7)^2 + (1)^2 - 25} = \sqrt{49 + 1 - 25} = \sqrt{50 - 25} = \sqrt{25} = 5$.

Let $\alpha$ be the angle between the two tangents drawn from the point $P_1 = (0, 0)$ to the circle.

The distance $d_1$ from $P_1$ to the center $C(7, -1)$ is:

$d_1 = \sqrt{(0-7)^2 + (0-(-1))^2} = \sqrt{(-7)^2 + (1)^2} = \sqrt{49 + 1} = \sqrt{50}$.

The angle $\theta$ between the tangents drawn from an external point to a circle is given by the formula $\sin(\theta/2) = \frac{r}{d}$, where $r$ is the radius of the circle and $d$ is the distance from the external point to the center of the circle.

For angle $\alpha$:

$\sin(\alpha/2) = \frac{r}{d_1} = \frac{5}{\sqrt{50}} = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}$.

Since $\sin(\alpha/2) = \frac{1}{\sqrt{2}}$, we have $\alpha/2 = \frac{\pi}{4}$.

Therefore, $\alpha = \frac{\pi}{2}$.

Let $\beta$ be the angle between the two tangents drawn from the point $P_2 = (8, 6)$ to the circle.

The distance $d_2$ from $P_2$ to the center $C(7, -1)$ is:

$d_2 = \sqrt{(8-7)^2 + (6-(-1))^2} = \sqrt{(1)^2 + (7)^2} = \sqrt{1 + 49} = \sqrt{50}$.

For angle $\beta$:

$\sin(\beta/2) = \frac{r}{d_2} = \frac{5}{\sqrt{50}} = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}$.

Since $\sin(\beta/2) = \frac{1}{\sqrt{2}}$, we have $\beta/2 = \frac{\pi}{4}$.

Therefore, $\beta = \frac{\pi}{2}$.

Finally, we need to find $\alpha - \beta$:

$\alpha - \beta = \frac{\pi}{2} - \frac{\pi}{2} = 0$.

The final answer is 0°.