Question

A circle drawn with the origin as the center passes through $(\frac{13}{2},0)$.

The point which does not lie in the interior of the circle is?

• A. $\frac{-3}{4},1$
• B. $2,\frac{7}{3}$
• C. $5\frac{-1}{2}$
• D. $(-6,\frac{5}{2})$

Answer 1

The correct option is D $(-6,\frac{5}{2})$

The circle passes through $(\frac{13}{2},0)$. Hence, radius of circle =$\sqrt{(6.5-0{)}^2+(0-0{)}^2}=6.5$

Hence, equation of circle is $x^2+y^2={6.5}^2$

If the sign of point when substituted in $x^2+y^2-{6.5}^2$ is positive then, it lies outside the circle.

Clearly, $(-\frac{3}{4},1)$ $({0.75}^2+1^2-{6.5}^2)$, $(2,\frac{7}{3})$ $(2^2+{2.33}^2-{6.5}^2)$ and $(5,-\frac{1}{2})$ $(5^2+{0.5}^2-{6.5}^2)$ have negative signs whereas $(-6,\frac{5}{2})$ $(6^2+{2.5}^2-{6.5}^2)$ has a positive sign.

Hence, this is the required answer.