AP EAMCET · Maths · Properties of Triangles
In \(\triangle A B C, \angle A=90^{\circ}\) and co-ordinates of points \(B\) and \(C\) are \((2,-4)\) and \((1,5)\). Then the equation of the circumcircle of \(\triangle A B C\) is
- A \(x^2+y^2+3 x+y+18=0\)
- B \(x^2+y^2-3 x+y-18=0\)
- C \(x^2+y^2-3 x-y-18=0\)
- D \(x^2+y^2+3 x-y+18=0\)
Answer & Solution
Correct Answer
(C) \(x^2+y^2-3 x-y-18=0\)
Step-by-step Solution
Detailed explanation
It is given that in \(\triangle A B C, \angle A=90^{\circ}\), so equation of circumcircle of \(\triangle A B C\), where \(B(2,-4)\) and \(C(1,5)\) \(\because B\) and \(C\) are end points of diameter of the circumcircle of \(\triangle A B C\), so equation of circumcircle is…
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