Question

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

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

Hint:

Angle between circles at intersection point is angle between their tangents, found using gradients \(\nabla S\). If acute angle comes, check supplementary angle as well.

Answer 1

The Correct Option is D

Step 1: Angle between circles equals angle between tangents at intersection.
This is equal to angle between their gradients (normals) at point of intersection.
Step 2: Write circle equations as \(S_1=0\) and \(S_2=0\).
\[ S_1=x^2+y^2-2x+8y+13 \]
\[ S_2=x^2+y^2-4x+6y+11 \]
Step 3: Find gradients.
\[ \nabla S_1=(2x-2,\;2y+8) \]
\[ \nabla S_2=(2x-4,\;2y+6) \]
At \((3,-4)\):
\[ \nabla S_1=(6-2,\;-8+8)=(4,0) \]
\[ \nabla S_2=(6-4,\;-8+6)=(2,-2) \]
Step 4: Find angle between normals.
\[ \cos\theta=\frac{\nabla S_1\cdot \nabla S_2}{|\nabla S_1||\nabla S_2|} \]
\[ = \frac{(4)(2)+(0)(-2)}{4\cdot \sqrt{(2)^2+(-2)^2}} = \frac{8}{4\cdot \sqrt{8}} = \frac{2}{2\sqrt{2}}=\frac{1}{\sqrt{2}} \]
Thus:
\[ \theta = 45^\circ \]
Angle between circles is supplementary angle:
\[ 180^\circ-45^\circ=135^\circ \]
Final Answer:
\[ \boxed{135^\circ} \]