TS EAMCET · Maths · Circle
If the lengths of tangents drawn to the circles \(x^2+y^2-8 x+40=0\) \(5 x^2+5 y^2-25 x+80=0\) \(x^2+y^2-8 x+16 y+160=0\) From the point \(P\) are equal, then \(P\) is equal to
- A \(\left(8, \frac{15}{2}\right)\)
- B \(\left(-8, \frac{15}{2}\right)\)
- C \(\left(8, \frac{-15}{2}\right)\)
- D \(\left(-8, \frac{-15}{2}\right)\)
Answer & Solution
Correct Answer
(C) \(\left(8, \frac{-15}{2}\right)\)
Step-by-step Solution
Detailed explanation
Let \(P\left(x_1, y_1\right)\) be the point from which the tangents are drawn to the circles \(S_1 \equiv x^2+y^2-8 x+40=0\) \(S_2 \equiv 5 x^2+5 y^2-25 x+80=0\) \(S_3 \equiv x^2+y^2-8 x+16 y+160=0\) Since, the length of the tangent from \(P\) to the circle \(S_1, S_2, S_3\) are…
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