Two circles each of radius 5 units touch each | vedclass

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(C) The common tangent is $4x+3y-10=0$. The slope of this tangent is $m = -\frac{4}{3}$.Since the line connecting the centers is perpendicular to the

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$x^2+y^2+6x+2y-15=0$

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(C) The common tangent is $4x+3y-10=0$. The slope of this tangent is $m = -\frac{4}{3}$.Since the line connecting the centers is perpendicular to the tangent,its slope is $m' = \frac{3}{4}$.Let the angle made by this line with the $x$-axis be $	heta$. Then $	an 	heta = \frac{3}{4}$,which gives $\sin 	heta = \frac{3}{5}$ and $\cos 	heta = \frac{4}{5}$.The centers of the circles are at a distance of $5$ units from the point of contact $(1,2)$ along the line perpendicular to the tangent.The coordinates of the centers are $(x,y) = (1 \pm 5 \cos 	heta, 2 \pm 5 \sin 	heta)$.$(x,y) = (1 \pm 5(\frac{4}{5}), 2 \pm 5(\frac{3}{5})) = (1 \pm 4, 2 \pm 3)$.Thus,the two possible centers are $C_1 = (1+4, 2+3) = (5,5)$ and $C_2 = (1-4, 2-3) = (-3,-1)$.The equations of the circles are $(x-5)^2 + (y-5)^2 = 25$ and $(x+3)^2 + (y+1)^2 = 25$.Expanding these,we get $x^2+y^2-10x-10y+25=0$ and $x^2+y^2+6x+2y-15=0$.$A$ circle passes through all four quadrants if its center $(h,k)$ satisfies $h^2 > r^2$ and $k^2 > r^2$.For $C_1(5,5)$,$5^2 = 25$,which is equal to $r^2$,so it touches the axes.For $C_2(-3,-1)$,$h^2 = (-3)^2 = 9 Thus,the required equation is $x^2+y^2+6x+2y-15=0$.

Two circles each of radius $5$ units touch each other at $(1,2)$ and $4x+3y=10$ is their common tangent. The equation of that circle among the two given circles,such that some portion of it lies in every quadrant is

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