JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
Let \(y=x+2,4 y=3 x+6\) and \(3 y=4 x+1\) be three tangent lines to the circle \((x-h)^2+(y-k)^2=r^2\). Then \(h+k\) is equal to :
- A \(5\)
- B \(5(1+\sqrt{2})\)
- C \(6\)
- D \(5 \sqrt{2}\)
Answer & Solution
Correct Answer
(A) \(5\)
Step-by-step Solution
Detailed explanation
\(L _1: y=x+2, L_2: 4 y=3 x+6, L_3: 3 y=4 x+1\) Bisector of lines \(L _2 \& L _3\) \(\frac{4 x-3 y+1}{5}=\pm\left(\frac{3 x-4 y+6}{5}\right)\) \(\text { (+) } 4 x-3 y+1=3 x-4 y+6\) \(x+y=5\) Centre lies on Bisector of \(4 x-3 y+1=0 \&\) \(\text { (0) } 3 x-4 y+6=0\)…
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