Question

The equation of circle which passes through the origin and cuts off intercepts $5$ and $6$ from the positive parts of the axes respectively, is

• A. ${(x+\frac{5}{2})}^2+{(y+3)}^2=\frac{61}{4}$
• B. ${(x-\frac{5}{2})}^2+{(y-3)}^2=\frac{61}{4}$
• C. ${(x+\frac{5}{2})}^2+{(y-3)}^2=\frac{61}{4}$
• D. ${(x-\frac{5}{2})}^2+{(y+3)}^2=\frac{61}{4}$

Answer 1

The correct option is B ${(x-\frac{5}{2})}^2+{(y-3)}^2=\frac{61}{4}$

From the figure, we have
$OP=5,OQ=6;OM=\frac{5}{2},CM=3$
In $△OMC$, we have
${OC}^2={OM}^2+{MC}^2$
$\Rightarrow {OC}^2={\left(\frac{5}{2}\right)}^2+3^2$
$\Rightarrow OC=\frac{\sqrt{61}}{2}$
Thus the required circle has its centre $(\frac{5}{2},3)$ and radius $\frac{\sqrt{61}}{2}$.
Hence, the equation of required circle is
${(x-\frac{5}{2})}^2+{(y-3)}^2=\frac{61}{4}$