Q76 / 180 · COMEDK 2021 · Mathematics
Find the centre and radius of the circle given by the equation $2 x^{2}+2 y^{2}+3 x+4 y+\dfrac{9}{8}=0$.
- Correct answerA1
- B$-1$
- C2
- D$-2$
Solution
The given equation of the circle is $2x^{2} + 2y^{2} + 3x + 4y + \dfrac{9}{8} = 0$.
Dividing the entire equation by $2$, we get $x^{2} + y^{2} + \dfrac{3}{2}x + 2y + \dfrac{9}{16} = 0$.
Comparing this with the standard form $x^{2} + y^{2} + 2gx + 2fy + c = 0$, we identify $2g = \dfrac{3}{2} \Rightarrow g = \dfrac{3}{4}$ and $2f = 2 \Rightarrow f = 1$.
The centre of the circle is $(-g, -f) = \left(-\dfrac{3}{4}, -1\right)$.
The radius $r$ is given by $\sqrt{g^{2} + f^{2} - c} = \sqrt{\left(\dfrac{3}{4}\right)^{2} + (1)^{2} - \dfrac{9}{16}} = \sqrt{\dfrac{9}{16} + 1 - \dfrac{9}{16}} = \sqrt{1} = 1$.
Answer: 1
Dividing the entire equation by $2$, we get $x^{2} + y^{2} + \dfrac{3}{2}x + 2y + \dfrac{9}{16} = 0$.
Comparing this with the standard form $x^{2} + y^{2} + 2gx + 2fy + c = 0$, we identify $2g = \dfrac{3}{2} \Rightarrow g = \dfrac{3}{4}$ and $2f = 2 \Rightarrow f = 1$.
The centre of the circle is $(-g, -f) = \left(-\dfrac{3}{4}, -1\right)$.
The radius $r$ is given by $\sqrt{g^{2} + f^{2} - c} = \sqrt{\left(\dfrac{3}{4}\right)^{2} + (1)^{2} - \dfrac{9}{16}} = \sqrt{\dfrac{9}{16} + 1 - \dfrac{9}{16}} = \sqrt{1} = 1$.
Answer: 1
Hence, the correct answer is option: 'A'.