The area (in sq. units) of the triangle forme | vedclass

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(B) The equation of the circle is $x^2+y^2-10x=0$. The center of the circle is $(5,0)$ and the radius is $5$.The equation of the tangent at point $P(9

Vedclass · Accepted Answer

$\frac{75}{8}$

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(B) The equation of the circle is $x^2+y^2-10x=0$. The center of the circle is $(5,0)$ and the radius is $5$.The equation of the tangent at point $P(9,3)$ is given by $xx_1 + yy_1 - 5(x+x_1) = 0$.Substituting $(x_1, y_1) = (9,3)$,we get $9x + 3y - 5(x+9) = 0$,which simplifies to $4x + 3y - 45 = 0$.The tangent intersects the $X$-axis $(y=0)$ at point $A$. Setting $y=0$ in the tangent equation: $4x = 45 \implies x = \frac{45}{4}$. So,$A = (\frac{45}{4}, 0)$.The normal at $(9,3)$ passes through the center $(5,0)$ and the point $(9,3)$. Its slope is $m = \frac{3-0}{9-5} = \frac{3}{4}$.The equation of the normal is $y - 0 = \frac{3}{4}(x - 5)$,which simplifies to $3x - 4y - 15 = 0$.The normal intersects the $X$-axis $(y=0)$ at point $B$. Setting $y=0$ in the normal equation: $3x = 15 \implies x = 5$. So,$B = (5, 0)$.The triangle is formed by vertices $P(9,3)$,$A(\frac{45}{4}, 0)$,and $B(5,0)$.The base of the triangle along the $X$-axis is $AB = |\frac{45}{4} - 5| = |\frac{45-20}{4}| = \frac{25}{4}$.The height of the triangle is the $y$-coordinate of $P$,which is $3$.Area $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes \frac{25}{4} 	imes 3 = \frac{75}{8}$ sq. units.

The area (in sq. units) of the triangle formed by the $X$-axis,the tangent,and the normal drawn to the circle $x^2+y^2=10x$ at the point $(9,3)$ is

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