TS EAMCET · Maths · Circle
Let a chord \(\mathrm{AB}\) subtend at angle of \(60^{\circ}\) at the centre \(\mathrm{C}(2,3)\) of a circle \(\mathrm{S}\). If the equation of \(\mathrm{AB}\) is \(\mathrm{x}+\mathrm{y}+1=0\), then the equation of the circle \(\mathrm{S}\) is
- A \(x^2+y^2-4 x-6 y+11=0\)
- B \(x^2+y^2-4 x-6 y+37=0\)
- C \(x^2+y^2-4 x-6 y-11=0\)
- D \(x^2+y^2-4 x-6 y-37=0\)
Answer & Solution
Correct Answer
(C) \(x^2+y^2-4 x-6 y-11=0\)
Step-by-step Solution
Detailed explanation
Length of perpendicular from \(C\) on \(A B\) \[ C D=\left|\frac{2+3+1}{\sqrt{1+1}}\right|=\frac{6}{\sqrt{2}}=3 \sqrt{2} \] In \(\triangle C A D\),…
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