Q64 / 180 · COMEDK 2024 (Memory Based) · Mathematics
The radius and centre of circle which equation is $3 x^2+3 y^2+6 x+12 y+5=0$ is
- A$(-3,-4), \sqrt{20}$
- B$(3,4), \sqrt{20}$
- Correct answerC$(-1,-2), 5 \sqrt{\frac{2}{3}}$
- D$(1,2), 5 \sqrt{\frac{2}{3}}$
Solution
If equation of circle is $x^2+y^2+2 g x+2 f g+c=0$
Then, centre $=(-g,-f)$
$\begin{aligned}<br/>& \text { Radius }=\sqrt{g^2+f^2-c} \\<br/>& \Rightarrow x^2+y^2+2 x+4 y+\frac{5}{3}=0<br/>\end{aligned}$
$\begin{aligned}<br/>& \therefore \text { Centre }=(-g,-f)=(-1,-2) \\<br/>& \therefore \text { Radius }=\sqrt{1^2+2^2-\frac{5}{3}}=\sqrt{5-\frac{5}{3}}=5 \sqrt{\frac{2}{3}}<br/>\end{aligned}$
Then, centre $=(-g,-f)$
$\begin{aligned}<br/>& \text { Radius }=\sqrt{g^2+f^2-c} \\<br/>& \Rightarrow x^2+y^2+2 x+4 y+\frac{5}{3}=0<br/>\end{aligned}$
$\begin{aligned}<br/>& \therefore \text { Centre }=(-g,-f)=(-1,-2) \\<br/>& \therefore \text { Radius }=\sqrt{1^2+2^2-\frac{5}{3}}=\sqrt{5-\frac{5}{3}}=5 \sqrt{\frac{2}{3}}<br/>\end{aligned}$
Hence, the correct answer is option: 'C'.