The distance from the centre of the circle $x^{2} + y^{2} = 2 x$ to the straight line passing through the points of intersection of the two circles. $x^{2} + y^{2} + 5 x - 8 y + 1 = 0$ & $x^{2} + y^{2} - 3 x + 7 y - 25 = 0$ is-

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Solution

The correct option is D $2$
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 : $C_{1} - C_{2}$
$x^{2} + y^{2} + 5 x - 8 y + 1 - x^{2} - y^{2} + 3 x - 7 y + 25 = 0$
$⟹ 8 x - 15 y + 26 = 0$
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, $d_{1} = \frac{a x_{2} + b y_{2} + c}{\sqrt{a^{2} + b^{2}}}$
$⟹ d_{1} = \frac{8 \times 1 - 15 \times 0 + 26}{\sqrt{(8)^{2} + (15)^{2}}}$
$⟹ d_{1} = \frac{34}{17} = 2$
Option C is the answer

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The distance from the centre of the circle x2+y2=2x to the straight line passing through the points of intersection of the two circles. x2+y2+5x−8y+1=0 & x2+y2−3x+7y−25=0 is-

The distance from the centre of the circle $x^{2} + y^{2} = 2 x$ to the straight line passing through the points of intersection of the two circles. $x^{2} + y^{2} + 5 x - 8 y + 1 = 0$ & $x^{2} + y^{2} - 3 x + 7 y - 25 = 0$ is-