Question

The point \( (3, -4) \) lies on both the circles \( x^{2} + y^{2} - 2x + 8y + 13 = 0 \) and \( x^{2} + y^{2} - 4x + 6y + 11 = 0 \). Then, the angle between the circles is

• A. $60^{\circ}$
• B. $\tan^{-1}(\frac{1}{2})$
• C. $\tan^{-1}(\frac{3}{5})$
• D. $135^{\circ}$

Hint:

Use $\cos \theta = \frac{d^{2}-r_{1}^{2}-r_{2}^{2}}{2r_{1}r_{2}}$ to find the angle between two intersecting circles.

Answer 1

The Correct Option is D

Step 1: Identify Centers and Radii
Circle 1: $C_{1}=(1,-4)$, $r_{1}=\sqrt{1^{2}+(-4)^{2}-13} = \sqrt{1+16-13} = 2$. Circle 2: $C_{2}=(2,-3)$, $r_{2}=\sqrt{2^{2}+(-3)^{2}-11} = \sqrt{4+9-11} = \sqrt{2}$.
Step 2: Distance between Centers
$d = C_{1}C_{2} = \sqrt{(2-1)^{2}+(-3-(-4))^{2}} = \sqrt{1^{2}+1^{2}} = \sqrt{2}$.
Step 3: Calculate the Angle
$\cos \theta = \frac{d^{2}-r_{1}^{2}-r_{2}^{2}}{2r_{1}r_{2}} = \frac{(\sqrt{2})^{2}-2^{2}-(\sqrt{2})^{2}}{2(2)(\sqrt{2})} = \frac{2-4-2}{4\sqrt{2}} = \frac{-4}{4\sqrt{2}} = -\frac{1}{\sqrt{2}}$. Thus, $\theta = 135^{\circ}$.
Final Answer: (d)