Question

$a x^{2}+2 y^{2}+2 b x y+2 x-y+c=0$ represents a circle through the origin, if
A .$\mathrm{a}=0, \mathrm{~b}=0, \mathrm{c}=2$
B. $a=1, b=0, c=0$
C. $a=2, b=2, c=0$
D. $a=2, b=0, c=0$

Answer 1

Hint:We are given an equation of a circle which passess through the origin. The equation for a circle has the generic form: ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$. By comparing the equation with the general equation, we find the values of a and b. As the circle passess through the origin so from that we also get the value of c.

Formula Used:
General form of a circle: ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$

Complete Step by step Solution:
We are aware that a second degree curve to be a circle satisfy the conditions of $a=b$ and $h=0$,
which gives us a $=2,b=0$.
Two dimensionally, the vertical y-axis and the horizontal x-axis are the two axes on the flat coordinate plane. Where they converge is where the origin lies. This point is typically designated with the letter O and has the coordinates (0,0).
We must also determine whether the origin meets the circle's equation or that $c=0$ in order for the circle to pass through the origin.
Then the circle passes through the origin if $a=2, b=0, c=0$

So the correct answer is option(D)

Note: Remember that the general form of the equation of a circle makes it difficult to identify any significant properties about any specific circle, in contrast to the standard form, which is simpler to comprehend. So, to quickly change from the generic form to the standard form, we will use the completing the square formula.The formula for a circle$\left(\mathrm{x}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}-\mathrm{y}_{1}\right)^{2}=\mathrm{r}^{2}$ whose centre is at $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$