- Home
- Standard 11
- Mathematics
10-1.Circle and System of Circles
hard
The straight line $x + 2y = 1$ meets the coordinate axes at $A$ and $B$. A circle is drawn through $A, B$ and the origin. Then the sum of perpendicular distances from $A$ and $B$ on the tangent to the circle at the origin is
A
$\frac {\sqrt 5}{2}$
B
$2\sqrt 5$
C
$\frac {\sqrt 5}{4}$
D
$4\sqrt 5$
(JEE MAIN-2019)
Solution
Equation of circle $x\left( {x – 1} \right) + \left( {y – \frac{1}{2}} \right)y = 0$
${x^2} + {y^2} – x – \frac{y}{2} = 0$
Equation of tangent at $\left( {0,0} \right)$
$x.0 + y.0 – \frac{{x + 0}}{2} – \frac{{y + 0}}{{2 \times 2}} = 0$
$2x + y = 0\,\,\,\,\,\,\,\,…….\left( 1 \right)$
Sum of distance of $A$ and $B$ from line $(i)$ is
$\frac{2}{{\sqrt 5 }} + \frac{{\frac{1}{2}}}{{\sqrt 5 }} = \frac{5}{{2\sqrt 5 }} = \frac{{\sqrt 5 }}{2}$
Standard 11
Mathematics
