### Find the radius, center and position of origin given an equation of circle with coefficient

Question Sample Titled 'Find the radius, center and position of origin given an equation of circle with coefficient '

The equation of the circle ${C}$ is ${2}{x}^{{2}}+{2}{y}^{{2}}+{32}{x}-{4}{y}-{365}={0}$. Which of the following are true?

 I. The radius of ${C}$ is ${25}$ . II. The origin lies inside ${C}$ . III. The coordinates of the centre of ${C}$ are ${\left(-{8},{1}\right)}$ .

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

 I, III.  ${2}{x}^{{2}}+{2}{y}^{{2}}+{32}{x}-{4}{y}-{365}$ $={0}$ ${x}^{{2}}+{y}^{{2}}+{16}{x}-{2}{y}-\dfrac{{365}}{{2}}$ $={0}$ Center $={\left(\dfrac{{{16}}}{{-{{2}}}},\dfrac{{-{2}}}{{-{{2}}}}\right)}={\left(-{8},{1}\right)}$ Radius $=\sqrt{{{\left(-{8}\right)}^{{2}}+{\left({1}\right)}^{{2}}+{365}}}$ $=\sqrt{{{430}}}$ I is false. III is true. II.  Subsitute origin ${\left({0},{0}\right)}$ into the left-handed side of the equation of ${C}$ . L.H.S. $={2}{\left({0}\right)}^{{2}}+{2}{\left({0}\right)}^{{2}}+{32}{\left({0}\right)}-{4}{\left({0}\right)}-{365}$ $=-{365}$ $\lt{0}$ ∴   The origin lies inside ${C}$ . undefinedundefined

 II. We can also compare the length of radius to the length of center to origin Length of origin to the center $=\sqrt{{{\left(-{8}-{0}\right)}^{{2}}+{\left({1}-{0}\right)}^{{2}}}}$ $=\sqrt{{{65}}}$ $\lt\sqrt{{{430}}}$ ∴   The origin lies inside ${C}$ . undefinedundefined

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