Consider a circle `x^2+y^2+a x+b y+c=0` lying completely in the first quadrant. If `m_1a n dm_2` are the maximum and minimum values of `y/x` for all ordered pairs `(x ,y)` on the circumference of the circle, then the value of `(m_1+m_2)` is `(a^2-4c)/(b^2-4c)` (b) `(2a b)/(b^2-4c)` `(2a b)/(4c-b^2)` (d) `(2a b)/(b^2-4a c)`

Options

A. `(a^(2)-4c)/(b^(2)-4c)`
B. `(2ab)/(b^(2)-4c)`
C. `(2ab)/(4c-b^(2))`
D. `(2ab)/(b^(2)-4ac)`

Answer

3

Explanation

Substituting `y=mx` in the equation of circle, we get
`x^(2)+m^(2)x^(2)+ax+bmx+c=0`
`(y//x` denotes the slope of the tangent from the origin on the circles. )
Since the line is touching the circle, we must have discriminant.
`(a+bm)^(2)-4c(1+m^(2))=0` or `a^(2)+b^(2)m^(2)+2abm-4c-4cm^(2)=0`
or `m^(2)(b^(2)-4c)+2abm+a^(2)-4c=0`
This equation has two roots `m_(1)` and `m_(2)`. Therefore, `m_(1)+m_(2)= -(2ab)/(b^(2)-4c)=(2ab)/(4c-b^(2))`