The perpendicular distance from the point (1, | vedclass

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(B) Given circles are $S_1: x^2+y^2-2x+4y-4=0$ and $S_2: x^2+y^2+4x-6y-3=0$. The equation of the common chord is given by $S_1 - S_2 = 0$. $(x^2+y^2-2

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$\frac{13}{\sqrt{136}}$

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(B) Given circles are $S_1: x^2+y^2-2x+4y-4=0$ and $S_2: x^2+y^2+4x-6y-3=0$. The equation of the common chord is given by $S_1 - S_2 = 0$. $(x^2+y^2-2x+4y-4) - (x^2+y^2+4x-6y-3) = 0$. $-6x + 10y - 1 = 0$,which simplifies to $6x - 10y + 1 = 0$. The perpendicular distance from the point $(x_1, y_1) = (1, 2)$ to the line $Ax + By + C = 0$ is given by $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$. Substituting the values: $d = \frac{|6(1) - 10(2) + 1|}{\sqrt{6^2 + (-10)^2}}$. $d = \frac{|6 - 20 + 1|}{\sqrt{36 + 100}} = \frac{|-13|}{\sqrt{136}} = \frac{13}{\sqrt{136}}$ units.

The perpendicular distance from the point $(1,2)$ to the common chord of the circles $x^2+y^2-2x+4y-4=0$ and $x^2+y^2+4x-6y-3=0$ is ........ units.

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