TS EAMCET · Maths · Circle
If the equations \(2 x-3 y+3=0,2 x+y+1=0\) and \(6 x+4 y+1=0\) represent the sides of a triangle, then the equation of the circle passing through the vertices of this triangle is
- A \(4 x^2+4 y^2+9 x-10 y+7=0\)
- B \(2 x^2+2 y^2-7 x-5 y+9=0\)
- C \(8 x^2+8 y^2+18 x-20 y+17=0\)
- D \(x^2+y^2+3 x-y+13=0\)
Answer & Solution
Correct Answer
(C) \(8 x^2+8 y^2+18 x-20 y+17=0\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text {} L_1: 2 x-3 y+3=0 \\ & L_2: 2 x+y+1=0 \\ & L_3: 6 x+4 y+1=0\end{aligned}\) \(\because \quad L_1 \perp L_3\) \(\therefore \quad\) Circle will have \(A B\) as diameter solving \(L_1, L_2\) and \(L_2, L_3\) for \(B\) and \(A\).…
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