TS EAMCET · Maths · Circle
If the length of the tangent from \((h, k)\) to the circle \(x^2+y^2=16\) is twice the length of the tangent from the same point to the circle \(x^2+y^2+2 x+2 y=0\), then
- A \(h^2+k^2+4 h+4 k+16=0\)
- B \(h^2+k^2+3 h+3 k=0\)
- C \(3 h^2+3 k^2+8 h+8 k+16=0\)
- D \(3 h^2+3 k^2+4 h+4 k+16=0\)
Answer & Solution
Correct Answer
(C) \(3 h^2+3 k^2+8 h+8 k+16=0\)
Step-by-step Solution
Detailed explanation
Given equations of circle is \[ x^2+y^2=16 \] and \(\quad x^2+y^2+2 x+2 y=0\) According to the questions, Length of the tangent from \((h, k)\) to the circle \[ x^2+y^2=16 \] \(=2 \times\) Length of the tangent from \((h, k)\) to the circle \(x^2+y^2+2 x+2 y=0\)…
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