Let the lengths of intercepts on the X-axis and Y-axis made by the circle x^2+y^2+ax+2ay+c…

Question

Let the lengths of intercepts on the $X$-axis and $Y$-axis made by the circle $x^2+y^2+ax+2ay+c=0$, ($a<0$), be $2\sqrt2$ and $2\sqrt5$, respectively. Then the shortest distance from the origin to a tangent to this circle which is perpendicular to the line $x+2y=0$, is equal to

Accepted Answer

$$2\sqrt {{{{a^2}} \over 4} - c}  = 2\sqrt 2 $$

$$\sqrt {{a^2} - 4c}  = 2\sqrt 2 $$

${a^2} - 4c = 8$ .... (1)

$$2\sqrt {{a^2} - c}  = 2\sqrt 5 $$

${a^2} - c = 5$ .... (2)

$(2) - (1)$

$$3c =  - 3a \Rightarrow c =  - 1$$

$${a^2} = 4 \Rightarrow a =  - 2$$  (Given a < 0)

Equation of circle

${x^2} + {y^2} - 2x - 4y - 1 = 0$

Equation of tangent which is perpendicular to the line x + 2y = 0 is

$$2x - y + \lambda  = 0$$

$$ \therefore $$ p = r

$$\left| {{{2 - 2 + \lambda } \over {\sqrt 5 }}} \right| = \sqrt 6 $$

$$ \Rightarrow \lambda  =  \pm \sqrt {30} $$

$$ \therefore $$ Tangent $$2x - y \pm \sqrt {30}  = 0$$

Distance from origin = $${{\sqrt {30} } \over {\sqrt 5 }} = \sqrt 6 $$

Solutions · 1