AP EAMCET · Maths · Circle
If the angle between the circles \(x^2+y^2-12 x-6 y+41=0\) and \(x^2+y^2+k x+6 y-59=0\) is \(45^{\circ}\), then a value of \(k\) is
- A 0
- B -4
- C -3
- D -1
Answer & Solution
Correct Answer
(B) -4
Step-by-step Solution
Detailed explanation
Given circles, \[ \begin{aligned} x^2+y^2-12 x-6 y+41 & =0 \\ \text { and } x^2+y^2+k x+6 y-59 & =0 \end{aligned} \] Centre of first circle is \((6,3)\) and radius \[ r_1=\sqrt{36-9-4}=2 \] Centre of second circle is \(\left(\frac{k}{2},-3\right)\) and radius…
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