TS EAMCET · Maths · Circle
The diameters of a circle are along \(2 x+y-7=0\) and \(x+3 y-11=0\). Then, the equation of this circle, which also passes through \((5,7)\), is
- A \(x^2+y^2-4 x-6 y-16=0\)
- B \(x^2+y^2-4 x-6 y-20=0\)
- C \(x^2+y^2-4 x-6 y-12=0\)
- D \(x^2+y^2+4 x+6 y-12=0\)
Answer & Solution
Correct Answer
(C) \(x^2+y^2-4 x-6 y-12=0\)
Step-by-step Solution
Detailed explanation
The intersection point of diameter lines is \((2,3)\) which is the centre of circle. Now, radius \(\begin{aligned} & =\sqrt{(5-2)^2+(7-3)^2} \\ & =\sqrt{9+16}=5 \end{aligned}\) \(\therefore\) Required equation of circle is…
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