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Find the equation of the circle with radius $5$ and contains the points $A(-8,0)$ and $B(-4,-2)$

Hint: Equate radii.

Thats the question from my midterms exam, can anyone help me finding this one, its a little tricky for me. Thanks, hope anyone can cite a good solution for me to follow. Thanks again..

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  • $\begingroup$ Does it pass through the two points or it contains them? $\endgroup$ Jan 28, 2014 at 6:57

3 Answers 3

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The centre of the circle lies on the perpendicular bisector of the line segment that joins $A$ and $B$.

The midpoint of this line segment is $(-6,-1)$. The line segment has slope $-\frac{1}{2}$, so the perpendicular bisector has slope $2$.

Now we can find the equation of the perpendicular bisector. It is $y=2x+11$.

The distance from $(x,2x+11)$ to $(-8,0)$ is $5$. Thus $$(x+8)^2+(2x+11)^2=25.$$ This simplifies to $$5x^2+60x+160=0,$$ and then to $x^2+12x+32=0$. The roots of this equation are easy to find: we get lucky, the quadratic even factors nicely.

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$$(8+x_1)^2 + y_1^2 = 5^2$$ $$(4+x_1)^2 + (2+y_1)^2 = 5^2$$ Subtracting the second equation from first: $$4(12+2x_1) - 2(2+2y_1) = 0$$ $$11 + 2x_1 = y_1$$

Substitute this in one of the two equations and find $x_1$ and then eventually $y_1$.Once you get the center of the circle $(x_1,y_1)$, you can calculate the equation of the circle using: $$(x-x_1)^2 + (y-y_1)^2 = r^2$$ where $r=5$ in this case.

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General eq. of circle is

$(x-a)^2+(y-b)^2=r^2$ where r is radius & $(a,b)$ is the center.

Hint:Put value of $r=5$ then

$(x-a)^2+(y-b)^2=25$

Now put points which are situated on circle in place of (x,y) one by one you get two eq. solve them for values of $a$ & $b$ then you get the eq. of circle by using these values in general eq.

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