Question

The equation of circle which passes through the origin and cuts off intercepts $5$ and $6$ from the positive parts of the axis respectively is ${(x-\frac{5}{2})}^2+(y-3{)}^2=\lambda$, where $\lambda$ is

• A. $\frac{61}{4}$
• B. $\frac{6}{4}$
• C. $\frac{1}{4}$
• D. $0$

Answer 1

Answer: (A)

$\frac{61}{4}$

From figure, we have
$OP=5,OQ=6$
and $OM=\frac{5}{2},CM=3$
Therefore, in $\Delta OMC,O{C}^2=O{M}^2+M{C}^2$
$\Rightarrow O{C}^2=O{M}^2+M{C}^2$
$\Rightarrow O{C}^2={\left(\frac{5}{2}\right)}^2+(3{)}^2$
$\Rightarrow OC=\frac{\overline{61}}{2}$
Thus, the requited circle has its centre $(\frac{5}{2},3)$ and radius $\frac{\overline{61}}{2}$.
Hence, its equation is ${(x-\frac{5}{2})}^2=(y-3{)}^2=\left(\frac{61}{2}\right)$
Hence, $\lambda =\frac{61}{4}$