# If a circle drawn with origin as the centre passes

Question:

If a circle drawn with origin as the centre passes through $\left(\frac{13}{2}, 0\right)$, then the point which does not lie in the interior of the circle is

(a) $\left(\frac{-3}{4}, 1\right)$

(b) $\left(2, \frac{7}{3}\right)$

(c) $\left(5, \frac{-1}{2}\right)$

(d) $\left(-6, \frac{5}{2}\right)$

Solution:

(d) it is given that, centre of circle in $(0,0)$ and passes through the point $\left(\frac{13}{2}, 0\right)$.

$\therefore$ Radius of circle $=$ Distance between $(0,0)$ and $\left(\frac{13}{2}, 0\right)$

$=\sqrt{\left(\frac{13}{2}-0\right)^{2}+(0-0)^{2}}=\sqrt{\left(\frac{13}{2}\right)^{2}}=\frac{13}{2}=6.5$

A point lie outside on or inside the circles of the distance of it from the centre of the circle is greater than equal to or less than radius of the circle.

Now, to get the correct option we have to check the option one by one.

(a) Distance between $(0,0)$ and $\left(\frac{-3}{4}, 1\right)=\sqrt{\left(\frac{-3}{4}-0\right)^{2}+(1-0)^{2}}$

$=\sqrt{\frac{9}{16}+1}=\sqrt{\frac{25}{16}}=\frac{5}{4}=1.25<6.5$

So, the point $\left(-\frac{3}{4}, 1\right)$ lies interior to the circle.

(b) Distance between $(0,0)$ and $\left(2, \frac{7}{3}\right)=\sqrt{(2-0)^{2}+\left(\frac{7}{3}-0\right)^{2}}$

$=\sqrt{4+\frac{49}{9}}=\sqrt{\frac{36+49}{9}}$

$=\sqrt{\frac{85}{9}}=\frac{9.22}{3}=3.1<6.5$

So, the point $\left(2, \frac{7}{3}\right)$ lies inside the circle.

(c) Distance between $(0,0)$ and $\left(5, \frac{-1}{2}\right)=\sqrt{(5-0)^{2}+\left(-\frac{1}{2}-0\right)^{2}}$

$=\sqrt{25+\frac{1}{4}}=\sqrt{\frac{101}{4}}=\frac{10.04}{2}$

$\Rightarrow$ $=5.02<6.5$

So, the point $\left(5,-\frac{1}{2}\right)$ lies inside the circle.

(d) Distance between $(0,0)$ and $\left(-6, \frac{5}{2}\right)=\sqrt{(-6-0)^{2}+\left(\frac{5}{2}-0\right)^{2}}$

$=\sqrt{36+\frac{25}{4}}=\sqrt{\frac{144+25}{4}}$

$=\sqrt{\frac{169}{4}}=\frac{13}{2}=6.5$

So, the point $\left(-6, \frac{5}{2}\right)$ lis an the circle i.e., does not lie interior to the circle.