AP EAMCET · Maths · Circle
The straight line touching the circle \(x^2+y^2-2 x-3=0\) and remaining normal to the circle \(x^2+y^2-4 y-6=0\) is
- A \(4 x-3 y+6=0\)
- B \(y+2=0\)
- C \(4 x+3 y-6=0\)
- D \(2 x+3=0\)
Answer & Solution
Correct Answer
(A) \(4 x-3 y+6=0\)
Step-by-step Solution
Detailed explanation
Line is normal to circle \(x^2+y^2-4 y-6=0\) \(\therefore\) It passes through centre \((0,2)\). Let \(m\) be slope of line. \(\therefore\) Equation of line is \(y-2=m x\) \[ \Rightarrow \quad m x-y+2=0 \] This line is tangent to \(x^2+y^2-2 x-3=0\) centre is \((1,0)\), radius is…
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