Question

Consider a circle $x^2+y^2+ax+by+c=0$ lying completely in the first quadrant. If $m_1$ and $m_2$ are the maximum and the minimum values of $\frac{y}{x}$ for all ordered pairs $(x,y)$ on the circumference of the circle, then the value of $(m_1+m_2)$ is

• A. $\frac{2ab}{4c-b^2}$
• B. $\frac{2ab}{b^2-4ac}$
• C. $\frac{2ab}{b^2-4c}$
• D. $\frac{2ab}{4ac-b^2}$

Answer 1

The correct option is A $\frac{2ab}{4c-b^2}$

Given $x^2+y^2+ax+by+c=0$
Let $m=\frac{y}{x}\Rightarrow y=mx$
Now, here $m=\frac{y}{x}$ represents slope of the line passing through $O(0,0)$ and a point on the circle.

Substituting $y=mx$ in$x^2+y^2+ax+by+c=0,$ we get
$x^2+m^2{x}^2+ax+bmx+c=0\Rightarrow (1+m^2)x^2+(a+bm)x+c=0$

Applying discriminant $=0$ ($\because$ For maximum and minimum value of $m,$ line $y=mx$ should be tangent to the circle)
$\Rightarrow (a+bm{)}^2-4(1+m^2)c=0$
$\Rightarrow m^2(b^2-4c)+2abm+a^2-4c=0$

So, we get two values of $m$, i.e. $m_1$ and $m_2,$ for maximum and minimum.
$\Rightarrow m_1+m_2=-\frac{2ab}{b^2-4c}=\frac{2ab}{4c-b^2}$