TS EAMCET · Maths · Circle
A circle S given by \(x^2+y^2-14 x+6 y+33=0\) cuts the X -axis at A and \(\mathrm{B}(\mathrm{OB}>\mathrm{OA})\). C is midpoint of \(\mathrm{AB}. \mathrm{L}\) is a line through C and having slope \((-1)\). If L is the diameter of a circle \(S^{\prime}\) and also the radical axis of the circles \(S\) and \(S^{\prime}\), then the equation of the circle \(S^{\prime}\) is
- A \(x^2+y^2-17 x+3 y+54=0\)
- B \(x^2+y^2+17 x-3 y-54=0\)
- C \(x^2+y^2-17 x+3 y+51=0\)
- D \(x^2+y^2-3 x+17 y-51=0\)
Answer & Solution
Correct Answer
(A) \(x^2+y^2-17 x+3 y+54=0\)
Step-by-step Solution
Detailed explanation
\(\text{For circle S, } y=0: x^2-14x+33=0\) \((x-3)(x-11)=0 \Rightarrow x=3, x=11\) \(\text{A}=(3,0), \text{B}=(11,0)\) \(\text{Midpoint C of AB}: C=\left(\frac{3+11}{2}, 0\right) = (7,0)\) \(\text{Line L through C}(7,0) \text{ with slope } -1: y-0=-1(x-7) \Rightarrow x+y-7=0\)…
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