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(C) Let the two given circles be $C_1: x^2 + y^2 + 5x - 8y + 1 = 0$ and $C_2: x^2 + y^2 - 3x + 7y - 25 = 0$.The equation of the common chord is given

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$2$

Vedclass · Answer

(C) Let the two given circles be $C_1: x^2 + y^2 + 5x - 8y + 1 = 0$ and $C_2: x^2 + y^2 - 3x + 7y - 25 = 0$.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$.The given circle is $x^2 + y^2 - 2x = 0$. Comparing this with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -1, f = 0$.The center of this circle is $(-g, -f) = (1, 0)$.The perpendicular distance $d$ from the point $(x_0, y_0) = (1, 0)$ 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: $d = \frac{|8(1) - 15(0) + 26|}{\sqrt{8^2 + (-15)^2}} = \frac{|8 + 26|}{\sqrt{64 + 225}} = \frac{34}{\sqrt{289}} = \frac{34}{17} = 2$.

Find the distance from the center of the circle $x^2 + y^2 = 2x$ to the common chord of the circles $x^2 + y^2 + 5x - 8y + 1 = 0$ and $x^2 + y^2 - 3x + 7y - 25 = 0$.

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