TS EAMCET · Maths · Circle
The centre of the smallest circle which cuts the circles \(x^2+y^2-2 x-4 y-4=0\) and \(x^2+y^2-10 x+12 y+52=0\) orthogonally is
- A \((1,2)\)
- B \((-3,2)\)
- C \((3,-2)\)
- D \((3,4)\)
Answer & Solution
Correct Answer
(C) \((3,-2)\)
Step-by-step Solution
Detailed explanation
Let the equation of the required circle be \(x^2+y^2+2 g x+2 f y+c=0\) and centre is \((-g,-f)\) Given circles, \(C_1: x^2+y^2-2 x-4 y-4=0\) Here, \(g_1=-1, f_1=-2, c_1=-4\) \(C_2: x^2+y^2-10 x+12 y+52=0\) Here, \(g_2=-5, f_2=6, c_2=52\) Condition of two circles cuts…
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