AP EAMCET · Maths · Straight Lines
The equation obtained by transforming \(x^2+y^2-6 x+10 y-2=0\) to the parallel axis through \((3,-5)\) is
- A \(x^2+y^2=16\)
- B \(x^2+y^2=9\)
- C \(x^2+y^2=25\)
- D \(x^2+y^2=36\)
Answer & Solution
Correct Answer
(D) \(x^2+y^2=36\)
Step-by-step Solution
Detailed explanation
Given equation is \[ \begin{aligned} x^2+y^2-6 x+10 y-2 & =0 \\ (x-3)^2+(y+5)^2 & =36 \end{aligned} \] The equation obtained by transforming \((x-3)^2+(y+5)^2=36\) to parallel axis through \((3,-5)\) is Put \(x-3=h\) and \(y+5=k\) \[ \therefore h^2+k^2=36 \] i.e. \(x^2+y^2=36\)
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