Find the equation of the circle which touches | vedclass

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(A) Let the center of the circle be $(h, k)$. Since the circle touches the $x$-axis and has a radius of $5$ in the first quadrant,the center must be $

Vedclass · Accepted Answer

$x^2 + y^2 - 30x - 10y + 225 = 0$

Vedclass · Answer

(A) Let the center of the circle be $(h, k)$. Since the circle touches the $x$-axis and has a radius of $5$ in the first quadrant,the center must be $(h, 5)$ where $h > 0$.The circle also touches the line $3x - 4y = 0$. The perpendicular distance from the center $(h, 5)$ to this line must be equal to the radius $5$.$\Rightarrow \left|\frac{3h - 4(5)}{\sqrt{3^2 + (-4)^2}}\right| = 5$$\Rightarrow \left|\frac{3h - 20}{5}\right| = 5$$\Rightarrow |3h - 20| = 25$This gives two cases: $3h - 20 = 25$ or $3h - 20 = -25$.Case $1$: $3h = 45 \Rightarrow h = 15$.Case $2$: $3h = -5 \Rightarrow h = -5/3$. Since the center is in the first quadrant,$h > 0$,so we take $h = 15$.The equation of the circle is $(x - 15)^2 + (y - 5)^2 = 5^2$.$x^2 - 30x + 225 + y^2 - 10y + 25 = 25$.$x^2 + y^2 - 30x - 10y + 225 = 0$.

Find the equation of the circle which touches the $x$-axis and the line $4y = 3x$,has its center in the first quadrant,and has a radius of $5$.

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