If $m_1$ and $m_2$ are the maximum and minimum values of $\frac{y}{x}$ for all ordered pairs $(x,y)$ on the circumference of the circle.

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 minimum values of $\frac{y}{x}$ for all ordered pairs $(x,y)$ on the circumference of the circle,then what is the value of $m_1+m_2?$

The equation of the circle $x^2+y^2+ax+by+c=0$ does not become a function of $\frac{y}{x}$ when i divide it by $x^2$.What should i do to solve this question?Please help me.

Answer 1

Given $x^2+y^2+ax+by+c=0$ and Let $\displaystyle m = \frac{y}{x}\Rightarrow y=mx$

Now here $\displaystyle m= \frac{y}{x}$ Represent Slope of line passes through $O(0,0)$

So Here for $\max$ and $\min$ of $m\;,$ means line $y=mx$ is tangent to the Circle

So Put $y=mx$ in $x^2+y^2+ax+by+c=0$ and Then Put $\bf{Discriminant = 0}$

So we get $x^2+m^2x^2+ax+bmx+c=0\Rightarrow (1+m^2)x^2+(a+bm)+c=0$

So $\displaystyle (a+bm)^2-4(1+m^2)\cdot c=0$

So we get two values of $m$ one for minimum and one for maximum.

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