The line 4x + 3y - 4 = 0 divides the circumfe | vedclass

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(A) The line divides the circumference in the ratio $1:2$,which corresponds to an arc angle of $\frac{1}{1+2} 	imes 360^{\circ} = 120^{\circ}$.Let $O

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$(x - 5)^2 + (y - 3)^2 = 10^2$

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(A) The line divides the circumference in the ratio $1:2$,which corresponds to an arc angle of $\frac{1}{1+2} 	imes 360^{\circ} = 120^{\circ}$.Let $O$ be the center $(5, 3)$ and $d$ be the perpendicular distance from $O$ to the line $4x + 3y - 4 = 0$.$d = \frac{|4(5) + 3(3) - 4|}{\sqrt{4^2 + 3^2}} = \frac{|20 + 9 - 4|}{5} = \frac{25}{5} = 5$.In the triangle formed by the center,the midpoint of the chord,and a point on the circumference,the angle at the center is half of the arc angle,i.e.,$60^{\circ}$.Using trigonometry,$\cos(60^{\circ}) = \frac{d}{R}$,where $R$ is the radius.$\frac{1}{2} = \frac{5}{R} \Rightarrow R = 10$.The equation of the circle with center $(5, 3)$ and radius $10$ is $(x - 5)^2 + (y - 3)^2 = 10^2$.

The line $4x + 3y - 4 = 0$ divides the circumference of a circle in the ratio $1:2$. If $C(5, 3)$ is the center of that circle,then the equation of the circle is

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