JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
Let a circle \(C\) touch the lines \(L_{1}: 4 x-3 y+K_{1}\) \(=0\) and \(L _{2}: 4 x -3 y + K _{2}=0, K _{1}, K _{2} \in R\). If a line passing through the centre of the circle \(C\) intersects \(L _{1}\) at \((-1,2)\) and \(L _{2}\) at \((3,-6)\), then the equation of the circle \(C\) is
- A \((x-1)^{2}+(y-2)^{2}=4\)
- B \((x+1)^{2}+(y-2)^{2}=4\)
- C \((x-1)^{2}+(y+2)^{2}=16\)
- D \((x-1)^{2}+(y-2)^{2}=16\)
Answer & Solution
Correct Answer
(C) \((x-1)^{2}+(y+2)^{2}=16\)
Step-by-step Solution
Detailed explanation
\(L _{1}: 4 x -3 y + K _{1}=0\) \(L _{2}: 4 x -3 y + K _{2}=0\) now \(-4-6+ K _{1}=0 \Rightarrow K _{1}=10\) \(12+18+ K _{2}=0 \Rightarrow K _{2}=-30\) \(\Rightarrow \quad\) Tangent to the circle are \(\quad 4 x -3 y +10=0\) \(\quad 4 x -3 y -30=0\) Length of diameter…
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