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A circle drawn with origin as the centre passes through $ \left(\frac{13}{2}, 0\right) $. 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) $ \left(-6, \frac{5}{2}\right) $
Given:
A circle drawn with origin as the centre passes through \( \left(\frac{13}{2}, 0\right) \).
To do:
We have to find the point which does not lie in the interior of the circle.
Solution:
Radius of the circle $=$ Distance between $(0,0)$ and $(\frac{13}{2}, 0)$
$=\sqrt{(\frac{13}{2}-0)^{2}+(0-0)^{2}}$
$=\sqrt{(\frac{13}{2})^{2}}$
$=\frac{13}{2}$
$=6.5$
A point lies outside on or inside the circle if the distance of it from the centre of the circle is greater than equal to or less than the radius of the circle respectively.
Therefore,
(a) The distance between \( (0,0) \) and \( \left(\frac{-3}{4}, 1\right)=\sqrt{\left(\frac{-3}{4}-0\right)^{2}+(1-0)^{2}} \)
$=\sqrt{\frac{9}{16}+1}$
$=\sqrt{\frac{25}{16}}$
$=\frac{5}{4}$
$=1.25<6.5$
This implies,
The point \( \left(-\frac{3}{4}, 1\right) \) lies inside the circle.
(b) The distance between \( (0,0) \) and \( \left(2, \frac{7}{3}\right)=\sqrt{(2-0)^{2}+\left(\frac{7}{3}-0\right)^{2}} \)
$=\sqrt{4+\frac{49}{9}}$
$=\sqrt{\frac{36+49}{9}}$
$=\sqrt{\frac{85}{9}}$
$=\frac{9.22}{3}$
$=3.1<6.5$
This implies,
The point \( \left(2, \frac{7}{3}\right) \) lies inside the circle.
(c) The distance between $(0,0)$ and $(5, \frac{-1}{2})=\sqrt{(5-0)^{2}+\left(-\frac{1}{2}-0\right)^{2}}$
$=\sqrt{25+\frac{1}{4}}$
$=\sqrt{\frac{101}{4}}$
$=\frac{10.04}{2}$
$=5.02<6.5$
This implies,
The point \( \left(5,-\frac{1}{2}\right) \) lies inside the circle.
(d) The distance between $(0,0)$ and $\left(-6, \frac{5}{2}\right) =\sqrt{(-6-0)^{2}+\left(\frac{5}{2}-0\right)^{2}}$
$=\sqrt{36+\frac{25}{4}}$
$=\sqrt{\frac{144+25}{4}}$
$=\sqrt{\frac{169}{4}}$
$=\frac{13}{2}$
$=6.5$
This implies,
The point \( \left(-6, \frac{5}{2}\right) \) lies outside the circle.
The point which does not lie in the interior of the circle is \( \left(-6, \frac{5}{2}\right) \).
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