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(B) Let the centre of the circle be $(h, k)$ and its radius be $r$. Since the circle touches the lines $4x-3y-24=0$ and $4x+3y-42=0$,the perpendicular

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$x^2+y^2-4x-6y-12=0$

Vedclass · Answer

(B) Let the centre of the circle be $(h, k)$ and its radius be $r$. Since the circle touches the lines $4x-3y-24=0$ and $4x+3y-42=0$,the perpendicular distance from $(h, k)$ to these lines is equal to $r$.$r = \left|\frac{4h-3k-24}{5}\right| = \left|\frac{4h+3k-42}{5}\right|$Also,the circle passes through $(2, 8)$,so $r^2 = (h-2)^2 + (k-8)^2$.From the distance equality,$4h-3k-24 = \pm(4h+3k-42)$.Case $1$: $4h-3k-24 = 4h+3k-42$ $\Rightarrow 6k = 18$ $\Rightarrow k = 3$.Substituting $k=3$ into the radius equation:$r^2 = \left(\frac{4h-3(3)-24}{5}\right)^2 = \left(\frac{4h-33}{5}\right)^2$Equating to $(h-2)^2 + (3-8)^2 = (h-2)^2 + 25$:$\frac{(4h-33)^2}{25} = (h-2)^2 + 25$$(4h-33)^2 = 25(h^2-4h+4+25) = 25(h^2-4h+29)$$16h^2 - 264h + 1089 = 25h^2 - 100h + 725$$9h^2 + 164h - 364 = 0$$(h-2)(9h+182) = 0$Since $h \le 8$,we take $h=2$. Thus,the centre is $(2, 3)$ and $r^2 = (2-2)^2 + (3-8)^2 = 25$.The equation is $(x-2)^2 + (y-3)^2 = 25 \Rightarrow x^2+y^2-4x-6y-12=0$.

The equation of a circle passing through the point $(2,8)$,touching the lines $4x-3y-24=0$ and $4x+3y-42=0$,and having the $x$-coordinate of its centre less than or equal to $8$ is

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