AP EAMCET · Maths · Circle
A circle passes through the centre of another circle \(x^2+y^2-3 x-4 y-1=0\) and whose centre is \((5,2)\). Then the equation of this circle is.........
- A \(4 x^2+4 y^2-40 x-16 y+67=0\)
- B \(3 x^2+3 y^2-40 x-16 y+67=0\)
- C \(2 x^2+2 y^2-40 x-16 y+67=0\)
- D \(x^2+y^2-10 x-4 y+67=0\)
Answer & Solution
Correct Answer
(A) \(4 x^2+4 y^2-40 x-16 y+67=0\)
Step-by-step Solution
Detailed explanation
The centre of the circle \(x^2+y^2-3 x-4 y-1=0\) is \(C(3 / 2,2)\), now as it is given that the required circle passes through the point \(C(3 / 2,2)\) and having centre \((5,2)\), so radius of required circle is \(r=\sqrt{\left(5-\frac{3}{2}\right)^2+(2-2)^2}=\frac{7}{2}\)…
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