MHT CET · Maths · Circle
The equation of the circle, the end-points of whose diameter are the centres of the circles \(x^{2}+y^{2}-2 x+3 y-3=0\) and \(x^{2}+y^{2}+6 x-12 y-5=0\) is
- A \(2 x^{2}+2 y^{2}+4 x-9 y-24=0\)
- B \(2 x^{2}+2 y^{2}+4 x+9 y-24=0\)
- C \(2 x^{2}+2 y^{2}+4 x-9 y+24=0\)
- D \(2 x^{2}+2 y^{2}-4 x-9 y-24=0\)
Answer & Solution
Correct Answer
(A) \(2 x^{2}+2 y^{2}+4 x-9 y-24=0\)
Step-by-step Solution
Detailed explanation
Since the quadrilateral \(A B C D\) is cyclic, we have
\(\mathrm{A}+\mathrm{C}=180^{\circ}\) and \(\mathrm{B}+\mathrm{D}=180^{\circ}\)
\(\therefore \cos \mathrm{A}=\cos \left(180^{\circ}-\mathrm{C}\right) \quad=-\cos \mathrm{C}\)
\(\cos \mathrm{B}=\cos \left(180^{\circ}-\mathrm{D}\right) \quad=-\cos \mathrm{D}\)
\(\therefore \cos \mathrm{A}+\cos \mathrm{B}+\cos \mathrm{C}+\cos \mathrm{D}=0\)
\(\mathrm{A}+\mathrm{C}=180^{\circ}\) and \(\mathrm{B}+\mathrm{D}=180^{\circ}\)
\(\therefore \cos \mathrm{A}=\cos \left(180^{\circ}-\mathrm{C}\right) \quad=-\cos \mathrm{C}\)
\(\cos \mathrm{B}=\cos \left(180^{\circ}-\mathrm{D}\right) \quad=-\cos \mathrm{D}\)
\(\therefore \cos \mathrm{A}+\cos \mathrm{B}+\cos \mathrm{C}+\cos \mathrm{D}=0\)
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