TS EAMCET · Maths · Circle
Each of the two orthogonal circles \(C_1\) and \(C_2\) passes through both the points \((2,0)\) and \((-2,0)\). If \(y=m x+c\) is a common tangent to these circles, then
- A \(c^2=4\left(1+2 m^2\right)\)
- B \(c^2=2\left(1+2 m^2\right)\)
- C \(c^2=1+m^2\)
- D \(c^2 m^2=4\left(1+m^2\right)\)
Answer & Solution
Correct Answer
(A) \(c^2=4\left(1+2 m^2\right)\)
Step-by-step Solution
Detailed explanation
Let the equation of the circle be \(x^2+y^2+2 g x+2 f y+c=0\) Since, this circle is passes through \((2,0)\) and \((-40)\) \[ \begin{aligned} \therefore \quad 4+4 g+c & =0, \\ 4-4 g+c & =0 \end{aligned} \] On solving Eqs. (i) and (ii), we get \(g=0, c=-4\) \(\therefore\)…
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