TS EAMCET · Maths · Circle
If the circle \(\mathrm{S}=0\) intersect the three circles \(\mathrm{S}_1 \equiv x^2+y^2+4 x-7=0\), \(\mathrm{S}_2 \equiv x^2+y^2+y=0\) and \(\mathrm{S}_3 \equiv x^2+y^2+\frac{3}{2} x+\frac{5}{2} y-\frac{9}{2}=0\) orthogonally, then radical axis of \(\mathrm{S}=0\) and \(\mathrm{S}_1=0\) is
- A \(4 x-y-7=0\)
- B \(x+y-3=0\)
- C \(4 x+y-3=0\)
- D \(x-y-2=0\)
Answer & Solution
Correct Answer
(C) \(4 x+y-3=0\)
Step-by-step Solution
Detailed explanation
For \(S \equiv x^2+y^2+2gx+2fy+c=0\): \(S_1 \equiv x^2+y^2+4x-7=0 \Rightarrow g_1=2, f_1=0, c_1=-7\) \(S_2 \equiv x^2+y^2+y=0 \Rightarrow g_2=0, f_2=1/2, c_2=0\) \(S_3 \equiv x^2+y^2+\frac{3}{2} x+\frac{5}{2} y-\frac{9}{2}=0 \Rightarrow g_3=3/4, f_3=5/4, c_3=-9/2\) Orthogonality…
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