AP EAMCET · Maths · Circle
The locus of the point of intersection of the tangents to the circle \(x^2+y^2=a^2\) which make complimentary angles with the \(X\)-axis is
- A \(x^2-y^2=0\)
- B \(x^2+y^2=0\)
- C \(x y=0\)
- D \(x^2+y^2=2 a^2\)
Answer & Solution
Correct Answer
(A) \(x^2-y^2=0\)
Step-by-step Solution
Detailed explanation
Let the point of intersection be \(P(x_1, y_1)\). Equation of pair of tangents: \((x^2+y^2-a^2)(x_1^2+y_1^2-a^2) = (xx_1+yy_1-a^2)^2\) Coefficient of \(x^2\), \(A = (x_1^2+y_1^2-a^2) - x_1^2 = y_1^2-a^2\) Coefficient of \(y^2\), \(B = (x_1^2+y_1^2-a^2) - y_1^2 = x_1^2-a^2\)…
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