The line 4x - 3y + 15 = 0 intersects the circ | vedclass

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(A) The equation of the circle is $x^2 + y^2 - 6x - 8y = 0$. The center is $(3, 4)$ and the radius $R = \sqrt{3^2 + 4^2} = 5$.The perpendicular distan

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(A) The equation of the circle is $x^2 + y^2 - 6x - 8y = 0$. The center is $(3, 4)$ and the radius $R = \sqrt{3^2 + 4^2} = 5$.The perpendicular distance $p$ from the center $(3, 4)$ to the line $4x - 3y + 15 = 0$ is $p = \frac{|4(3) - 3(4) + 15|}{\sqrt{4^2 + (-3)^2}} = \frac{|12 - 12 + 15|}{5} = \frac{15}{5} = 3$.The length of the chord $AB = 2 \sqrt{R^2 - p^2} = 2 \sqrt{5^2 - 3^2} = 2 \sqrt{25 - 9} = 2 	imes 4 = 8$.The area of $\Delta ABC$ is maximum when the height of the triangle is maximum. The height is maximum when $C$ is at the end of the diameter perpendicular to the chord $AB$. The maximum height $h = R + p = 5 + 3 = 8$.Therefore,the maximum area $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 8 	imes 8 = 32 \ sq. \ units$.

The line $4x - 3y + 15 = 0$ intersects the circle $x^2 + y^2 - 6x - 8y = 0$ at two points $A$ and $B$. The maximum area of $\Delta ABC$,where $C$ is a point on the circumference of the circle,will be - .............. $sq. \ units$.

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