Question

In the $xy$ plane, the segment with end points$(3,8)$ and $(5,2)$ is the diameter of the circle. The point $(k,10)$ lies on the circle for:

• A. no value of $k$
• B. exactly one integral $k$
• C. exactly one non integral $k$
• D. two real values of $k$

Answer 1

The correct option is A no value of $k$

The centre of the given circle is,

$C(x,y)=(\frac{3+5}{2},\frac{8+2}{2})=(4,5)$

The length of the diameter of the circle is,

$D=\sqrt{{(5-3)}^2+{(2-8)}^2}=2\sqrt{10}units$

Therefore,

Radius $=\sqrt{10}units$

Therefore, the equation of the required circle is,

${(x-4)}^2+{(y-5)}^2=10$

Since, the point $(k,10)$ lies on the circle, we have

${(k-4)}^2+{(10-5)}^2=10$

$k^2-8k+16+25=10$

$k^2-8k+31=0$

This equation has no real roots. Therefore, no such value of $k$ exists.