A circle passes through (-2, 4) and touches t | vedclass

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Question

(A) Let the equation of the circle be $(x - h)^2 + (y - k)^2 = r^2.$Since it touches the $y-$axis at $(0, 2)$,the center is $(r, 2)$ or $(-r, 2)$.Sinc

Vedclass · Accepted Answer

$2x - 3y + 10 = 0$

Vedclass · Answer

(A) Let the equation of the circle be $(x - h)^2 + (y - k)^2 = r^2.$Since it touches the $y-$axis at $(0, 2)$,the center is $(r, 2)$ or $(-r, 2)$.Since the circle passes through $(-2, 4)$,the center must be at $(-r, 2)$.The radius $r$ is the distance from the center $(-r, 2)$ to the point $(0, 2)$,which is $r$.The distance from the center $(-r, 2)$ to $(-2, 4)$ is also $r$.So,$(-r - (-2))^2 + (2 - 4)^2 = r^2$$(2 - r)^2 + (-2)^2 = r^2$$4 - 4r + r^2 + 4 = r^2$$8 - 4r = 0 \Rightarrow r = 2.$Thus,the center is $(-2, 2).$$A$ diameter of the circle must pass through the center $(-2, 2).$Checking the options:For $A: 2(-2) - 3(2) + 10 = -4 - 6 + 10 = 0.$Since the center $(-2, 2)$ satisfies the equation $2x - 3y + 10 = 0$,this line is a diameter.

$A$ circle passes through $(-2, 4)$ and touches the $y-$axis at $(0, 2).$ Which one of the following equations can represent a diameter of this circle?

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