Question

The smallest circle with center on $y$-axis and passing through the point $\left(7,3\right)$ has radius

• A. $\sqrt{58}$
• B. $7$
• C. $3$
• D. $4$

Answer 1

The correct option is B

$7$

Explanation for the correct answer:

Compute the radius:

Let the coordinates of the center be $\left(h,k\right)$

Since the center is on the $y$ axis , $h=0$

Hence, the coordinates if the center are $\left(0,k\right)$

As point $\left(7,3\right)$ lies on the circle, the distance between it and the center must be equal to the radius.

$\therefore r^2={\left(7-0\right)}^2+{\left(3-k\right)}^2$

$\Rightarrow$ $r^2=49+9+k^2-6k$

$\Rightarrow$ $r^2=k^2-6k+58$

To find the minimum value we must differentiate with respect to $k$ and equate it to $0$

$2r\frac{dr}{dk}=2k-6$

$\Rightarrow$ $\frac{dr}{dk}=\frac{k-3}{r}$

$\Rightarrow$ $\frac{k-3}{r}=0$

$\Rightarrow$ $k=3$

Hence,

$r^2=3^2-6\times 3+58$

$\Rightarrow$$r^2=49$

$\Rightarrow$ $r=7$

Hence, the radius of the smallest circle satisfying the given conditions is $7$.

Hence, option $\left(B\right)$ is the correct answer.