### Determine whether a point is outside a circle and the angle given an equation of circle

Question Sample Titled 'Determine whether a point is outside a circle and the angle given an equation of circle'

The equation of the circle ${C}$ is ${3}{x}^{{2}}+{3}{y}^{{2}}+{27}{x}-{81}{y}-\dfrac{{53}}{{2}}={0}$ . The coodinates of the points ${P}$ and ${Q}$ are ${\left(-{20},{4}\right)}$ and ${\left({4},-{10}\right)}$ respectively. Which of the following is/are true?

 I. The radius of ${C}$ is ${43}$ . II. The mid-point of ${P}{Q}$ lies outside ${C}$ . III. If ${G}$ is the centre of ${C}$, then $\angle{P}{G}{Q}$ is an acute angle.

A
II and III only
B
III only
C
II only
D
I and III only

 I. ${3}{x}^{{2}}+{3}{y}^{{2}}+{27}{x}-{81}{y}-\dfrac{{53}}{{2}}$ $={0}$ ${x}^{{2}}+{y}^{{2}}+{9}{x}-{27}{y}-\dfrac{{53}}{{6}}$ $={0}$ Centre of ${C}$ $={\left(\dfrac{{{9}}}{{-{{2}}}},\dfrac{{-{27}}}{{-{{2}}}}\right)}={\left(-\dfrac{{9}}{{2}},\dfrac{{27}}{{2}}\right)}$ Radius of ${C}$ $=\sqrt{{{\left(-\dfrac{{9}}{{2}}\right)}^{{2}}+{\left(\dfrac{{27}}{{2}}\right)}^{{2}}-{\left(-\dfrac{{53}}{{6}}\right)}}}=\dfrac{{\sqrt{{{1902}}}}}{{3}}$ I is false. II. Let ${M}$ be the mid-point of ${P}{Q}$ . ${M}$ $={\left(\dfrac{{-{20}+{4}}}{{2}},\dfrac{{{4}-{10}}}{{2}}\right)}={\left(-{8},-{3}\right)}$ Let ${G}$ be the centre of ${C}$ . ${G}$ $={\left(-\dfrac{{9}}{{2}},\dfrac{{27}}{{2}}\right)}$ . Compare the length of ${G}{M}$ and the radius to determine whether ${M}$ lies inside or outside ${C}$ . ${G}{M}$ $=\sqrt{{{\left(-{8}+\dfrac{{9}}{{2}}\right)}^{{2}}+{\left(-{3}-\dfrac{{27}}{{2}}\right)}^{{2}}}}=\dfrac{{\sqrt{{{1138}}}}}{{2}}\approx{16.867}$ Radius  $=\dfrac{{\sqrt{{{1902}}}}}{{3}}\approx{14.537}$ ∴  ${M}$ lies outside ${C}$ . II is true. III. ${P}{Q}$ $=\sqrt{{{\left(-{20}-{4}\right)}^{{2}}+{\left({4}+{10}\right)}^{{2}}}}={2}\sqrt{{{193}}}$ ${P}{G}$ $=\sqrt{{{\left(-{20}+{4.5}\right)}^{{2}}+{\left({4}-{13.5}\right)}^{{2}}}}=\dfrac{{\sqrt{{{1322}}}}}{{2}}$ ${Q}{G}$ $=\sqrt{{{\left({4}+{4.5}\right)}^{{2}}+{\left(-{10}-{13.5}\right)}^{{2}}}}=\dfrac{{\sqrt{{{2498}}}}}{{2}}$ ${\cos}\angle{P}{G}{Q}$ $=\dfrac{{{\left({P}{G}\right)}^{{2}}+{\left({Q}{G}\right)}^{{2}}-{\left({P}{Q}\right)}^{{2}}}}{{{2}\cdot{P}{G}\cdot{Q}{G}}}=\dfrac{{{\left(\dfrac{{\sqrt{{{1322}}}}}{{2}}\right)}^{{2}}+{\left(\dfrac{{\sqrt{{{2498}}}}}{{2}}\right)}^{{2}}-{\left({2}\sqrt{{{193}}}\right)}^{{2}}}}{{{2}\cdot\dfrac{{\sqrt{{{1322}}}}}{{2}}\cdot\dfrac{{\sqrt{{{2498}}}}}{{2}}}}\approx{0.201}$ $\angle{P}{G}{Q}$ $\approx{78.4}^{\circ}$ ∴  $\angle{P}{G}{Q}$ is an acute angle. III is true.

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