MHT CET · Maths · Circle
The equation of a circle passing through origin and making \(\mathrm{x}\) -intercept 3 and \(\mathrm{y}\) - intercept \(-5\) is
- A \(x^{2}+y^{2}+3 x+5 y=0\)
- B \(x^{2}+y^{2}+3 x-5 y=0\)
- C \(x^{2}+y^{2}-3 x+5 y=0\)
- D \(x^{2}+y^{2}-3 x-5 y=0\)
Answer & Solution
Correct Answer
(C) \(x^{2}+y^{2}-3 x+5 y=0\)
Step-by-step Solution
Detailed explanation
\(\because A(3,0), B(0,-5)\) be the co-ordinate of ends of diameter AB. By diameter form, equation of circle is
\(
\begin{array}{l}
(x-3)(x-0)+(y-0)(y+5)=0 \\
x^{2}-3 x+y^{2}+5 y=0 \Rightarrow x^{2}+y^{2}-3 x+5 y=0
\end{array}
\)
\(
\begin{array}{l}
(x-3)(x-0)+(y-0)(y+5)=0 \\
x^{2}-3 x+y^{2}+5 y=0 \Rightarrow x^{2}+y^{2}-3 x+5 y=0
\end{array}
\)
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