The locus of the centers of the circles touch | vedclass

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(C) Let the center of the circle be $(h, k)$. Since the circle touches both lines,the perpendicular distance from $(h, k)$ to both lines must be equal

Vedclass · Accepted Answer

Both $(A)$ and $(B)$

Vedclass · Answer

(C) Let the center of the circle be $(h, k)$. Since the circle touches both lines,the perpendicular distance from $(h, k)$ to both lines must be equal to the radius $r$.So,$\frac{|3h - 4k + 1|}{\sqrt{3^2 + (-4)^2}} = \frac{|12h + 5k - 1|}{\sqrt{12^2 + 5^2}}$.This simplifies to $\frac{|3h - 4k + 1|}{5} = \frac{|12h + 5k - 1|}{13}$.Taking the positive sign: $13(3h - 4k + 1) = 5(12h + 5k - 1) \implies 39h - 52k + 13 = 60h + 25k - 5 \implies 21h + 77k - 18 = 0$.Taking the negative sign: $13(3h - 4k + 1) = -5(12h + 5k - 1) \implies 39h - 52k + 13 = -60h - 25k + 5 \implies 99h - 27k + 8 = 0$.Replacing $(h, k)$ with $(x, y)$,the locus is $21x + 77y - 18 = 0$ and $99x - 27y + 8 = 0$.

The locus of the centers of the circles touching the lines $3x - 4y + 1 = 0$ and $12x + 5y - 1 = 0$ is/are:

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