TS EAMCET · Maths · Circle
The radius of a circle \(C_1\) is thrice the radius of another circle \(C_2\) and the centres of \(C_1\) and \(C_2\) are \((1,2)\) and \((3,-2)\) respectively. If they cut each other orthogonally and the radius of the circle \(\mathrm{C}_1\) is 3 r, then the equation of the circle with r as radius and \((1,-2)\) as centre is
- A \(x^2+y^2-2 x+4 y-3=0\)
- B \(x^2+y^2-2 x+4 y+7=0\)
- C \(x^2+y^2-2 x+4 y-7=0\)
- D \(x^2+y^2-2 x+4 y+3=0\)
Answer & Solution
Correct Answer
(D) \(x^2+y^2-2 x+4 y+3=0\)
Step-by-step Solution
Detailed explanation
\(O_1=(1,2)\), \(R_1=3r\) \(O_2=(3,-2)\), \(R_2=r\) \(d^2 = (3-1)^2 + (-2-2)^2 = 2^2 + (-4)^2 = 4+16=20\) \(R_1^2 + R_2^2 = d^2\) \((3r)^2 + r^2 = 20\) \(9r^2 + r^2 = 20\) \(10r^2 = 20\) \(r^2 = 2\) Equation of circle with center \((1,-2)\) and radius \(r\):…
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