TS EAMCET · Maths · Circle
If \(5 x+6 y-34=0\) and \(2 x+y+c=0\) are conjugate lines with respect to the circle \(x^2+y^2-8 x-10 y+25=0\) then the point on the line \(2 x+y+c=0\) is
- A \((3,3)\)
- B \((2,4)\)
- C \((1,-5)\)
- D \((-2,-2)\)
Answer & Solution
Correct Answer
(C) \((1,-5)\)
Step-by-step Solution
Detailed explanation
Given lines are \(5 x+6 y-34=0\) and \(2 x+y+C=0\) with respect to \(x^2+y^2-8 x-10 y+25=0\). \[ \begin{aligned} & x^2+y^2-8 x-10 y+25+16-16=0 \\ & \left(x^2-8 x+16\right)+\left(y^2-10 y+25\right)=16 \\ & (x-4)^2+(y-5)^2=(4)^2 \end{aligned} \] Here, centre is \((4,5)\) and…
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