- Home
- Standard 11
- Mathematics
The distance from the centre of the circle $x^2 + y^2 = 2x$ to the straight line passing through the points of intersection of the two circles $x^2 + y^2 + 5x -8y + 1 =0$ and $x^2 + y^2-3x + 7y -25 = 0$ is-
$1$
$3$
$2$
$1/3$
Solution
Let $C_{1}, C_{2}$ be the two intersecting circles. Equation of $C_{1}: x^{2}+y^{2}+5 x-8 y+1=0$
Equation of $C_{2}: x^{2}+y^{2}-3 x+7 y-25=0$
Equation of the common chord of the two intersecting circles : $C_{1}-C_{2}$
$x^{2}+y^{2}+5 x-8 y+1-x^{2}-y^{2}+3 x-7 y+25=0$
$\Longrightarrow 8 x-15 y+26=0$
For the given circle with eqn $x^{2}+y^{2}=2 x$
Centre of the above circle $=(1,0)$
To find the distance between the centre and a line, $d_{1}=\frac{a x_{2}+b y_{2}+c}{\sqrt{a^{2}+b^{2}}}$
$\Longrightarrow d_{1}=\frac{8 \times 1-15 \times 0+26}{\sqrt{(8)^{2}+(15)^{2}}}$
$\Longrightarrow d_{1}=\frac{34}{17}=2$
Option $\mathrm{C}$ is the answer
