The distance from the centre of the circle x^ | vedclass

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(C) Let $C_{1}$ and $C_{2}$ be the two intersecting circles.Equation of $C_{1}: x^{2} + y^{2} + 5x - 8y + 1 = 0$Equation of $C_{2}: x^{2} + y^{2} - 3x

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) Let $C_{1}$ and $C_{2}$ be the two intersecting circles.Equation of $C_{1}: x^{2} + y^{2} + 5x - 8y + 1 = 0$Equation of $C_{2}: x^{2} + y^{2} - 3x + 7y - 25 = 0$The equation of the common chord is given by $C_{1} - C_{2} = 0$.$(x^{2} + y^{2} + 5x - 8y + 1) - (x^{2} + y^{2} - 3x + 7y - 25) = 0$$8x - 15y + 26 = 0$For the circle $x^{2} + y^{2} - 2x = 0$,the centre is $(1, 0)$.The distance $d$ from the point $(x_{0}, y_{0})$ to the line $ax + by + c = 0$ is given by $d = \frac{|ax_{0} + by_{0} + c|}{\sqrt{a^{2} + b^{2}}}$.Substituting the values: $d = \frac{|8(1) - 15(0) + 26|}{\sqrt{8^{2} + (-15)^{2}}}$$d = \frac{|8 + 26|}{\sqrt{64 + 225}} = \frac{34}{\sqrt{289}} = \frac{34}{17} = 2$.Thus,the distance is $2$.

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-

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