Question

Question 16
If a circle drawn with origin as the centre, passes through $(\frac{13}{2},0)$, then 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

It is given that, centre of circle is (0,0) and it passes through the point $(\frac{13}{2},0)$.
$\therefore$ Radius of circle = Distance between (0,0) and $(\frac{13}{2},0)$
$=\sqrt{{(\frac{13}{2}-0)}^2+(0-0{)}^2}=\sqrt{{\left(\frac{13}{2}\right)}^2}=\frac{13}{2}=6.5$
A point lie outside, on or inside the circle if the distance of point from the centre of the circle is greater than, equal to or less than radius of the circle respectively.
Now, to get the correct option, we have to check the option one by one.
(a) Distance between (0,0) and $(\frac{-3}{4},1)$
$=\sqrt{(\frac{-3}{4}-0{)}^2+(1-0{)}^2}$
$=\sqrt{\frac{9}{16}+1}$$=\sqrt{\frac{25}{16}}$
$=\frac{5}{4}$=$1.25$ <$6.5$
So, the point $(\frac{-3}{4},1)$ lies in the interior of the circle.
(b) Distance between (0,0) and $(2,\frac{7}{3})=$
$\sqrt{(2-0{)}^2+{(\frac{7}{3}-0)}^2}$
$\sqrt{4+\frac{49}{9}}=\sqrt{\frac{36+49}{9}}=\sqrt{\frac{85}{9}}=\frac{9.22}{3}=3.1<6.5$
So, the point $(2,\frac{7}{3})$ lies in the interior of the circle.
(c) Distance between (0,0) and $(5,\frac{-1}{2})$
$=\sqrt{(5-0{)}^2+{(-\frac{1}{2}-0)}^2}=\sqrt{25+\frac{1}{4}}=\sqrt{\frac{101}{4}}=\frac{10.04}{2}\Rightarrow 5.02<6.5$)
$So,thepoint(5,\frac{-1}{2})$ lies in the interior of the circle.
(d) Distance between (0,0) and $(-6,\frac{5}{2})$
$=\sqrt{(-6-0{)}^2+{(\frac{5}{2}-0)}^2}=\sqrt{36+\frac{25}{4}}=\sqrt{\frac{144+25}{4}}=\sqrt{\frac{169}{4}}=\frac{13}{2}=6.5$
So, the point $(-6,\frac{5}{2})$ lies on the circle i.e., does not lie in the interior of the circle.