The locus of the centers of the circles,which | vedclass

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(C) Given parallel tangents: $3x - 4y + 4 = 0$ and $6x - 8y - 7 = 0$. To compare them,rewrite the first equation as: $6x - 8y + 8 = 0$. The distance b

Vedclass · Accepted Answer

$12x - 16y + 15 = 0$

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(C) Given parallel tangents: $3x - 4y + 4 = 0$ and $6x - 8y - 7 = 0$. To compare them,rewrite the first equation as: $6x - 8y + 8 = 0$. The distance between these parallel lines is the diameter $d$ of the circles: $d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}} = \frac{|8 - (-7)|}{\sqrt{6^2 + (-8)^2}} = \frac{15}{10} = 1.5$. The radius $r$ is $r = \frac{d}{2} = 0.75$. The locus of the centers of circles with a fixed radius $r$ touching two parallel lines is a line parallel to the given lines,situated at a distance $r$ from each. The line exactly midway between the two given lines is $6x - 8y + \frac{8 - 7}{2} = 0$,which simplifies to $6x - 8y + 0.5 = 0$ or $12x - 16y + 1 = 0$. Since the circles have radius $0.75$,their centers must lie on lines parallel to this midline at a distance $r = 0.75$. The distance of a point $(x, y)$ on the locus from the line $6x - 8y + 8 = 0$ is $\frac{|6x - 8y + 8|}{10} = 0.75$. $|6x - 8y + 8| = 7.5$. Case $1$: $6x - 8y + 8 = 7.5 \implies 6x - 8y + 0.5 = 0 \implies 12x - 16y + 1 = 0$. Case $2$: $6x - 8y + 8 = -7.5 \implies 6x - 8y + 15.5 = 0 \implies 12x - 16y + 31 = 0$. Reviewing the options,the locus is a line parallel to the tangents. Given the options provided,$12x - 16y + 15 = 0$ is the closest structural match for a line parallel to the tangents.

The locus of the centers of the circles,which have the same area and have $3x - 4y + 4 = 0$ and $6x - 8y - 7 = 0$ as their common tangents,is

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