Question

Find the center and radius of the circle who's equation is 2x^2+2y^2-3x+2y+1

Answer 1

Answer:

Step-by-step explanation:

We have,

The equation of the circle

$\tt{S:2x^2+2y^2-3x+2y+1=0}$

$\tt{\implies\,S:\,x^2+y^2-\dfrac{3}{2}x+y+\dfrac{1}{2}=0}$

$\sf{Here,\,\,\,g=-\dfrac{3}{4}\,\,\,\,\,\,\,\,and\,\,\,\,\,\,\,\,f=\dfrac{1}{2}}$

So,

$\sf{\blue{Centre\equiv\bigg(\dfrac{3}{4},-\dfrac{1}{2}\bigg)}}$

Now,

$\sf{Radius=\,\sqrt{\bigg(-\dfrac{3}{4}\bigg)^2+\bigg(\dfrac{1}{2}\bigg)^2-\dfrac{1}{2}}}$

$\sf{=\sqrt{\dfrac{9}{16}+\dfrac{1}{4}-\dfrac{1}{2}}}$

$\sf{=\sqrt{\dfrac{9}{16}-\dfrac{1}{4}}}$

$\sf{=\sqrt{\dfrac{9-4}{16}}}$

$\sf{=\sqrt{\dfrac{5}{16}}}$

$\sf{=\dfrac{\sqrt{5}}{4}}$