TS EAMCET · Maths · Circle
The tangent at \(A(-1,2)\) on the circle \(x^2+y^2-4 x-8 y+7=0\) touches the circle \(x^2+y^2+4 x+6 y=0\) at \(B\). Then, a point of trisection of \(A B\) is
- A \(\left(0, \frac{1}{3}\right)\)
- B \(\left(-\frac{1}{3}, 1\right)\)
- C \(\left(\frac{2}{3}, \frac{1}{3}\right)\)
- D \((-1,-1)\)
Answer & Solution
Correct Answer
(B) \(\left(-\frac{1}{3}, 1\right)\)
Step-by-step Solution
Detailed explanation
Let circle, \(C_1=x^2+y^2-4 x-8 y+7=0\) Centre \((2,4)\) Radius \(r=\sqrt{13}\) Circle \(C_2=x^2+y^2+4 x+6 y=0\) Centre: \((-2,-3)\) Radius \(r=\sqrt{13}\) Radius of circle \(C_1\) and \(C_2\) are equal coordinate of \(B\) is mid-point of centre of circle \(C_1\) and \(C_2\).…
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