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# help circles

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158
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The equation of a circle which has a center at (-5, 2) can be written as Ax^2 + 2y^2 + Bx + Cy = $$80$$. Find the radius of the circle

Apr 15, 2022

#1
+66
-1

Since the center is at (-5, 2), The equation of the circle is $$(x+5)^2+(y-2)^2=r^2$$

Then you can multiply out $$x^2+10x+25+y^2-4y+4=r^2$$

$$x^2+y^2+10x-4y+29=r^2 \\ 2x^2+2y^2+20x-8y=2r^2-58 \\ 2r^2-58=80 \\ r^2-29=40 \\ r^2=69 \\ r=\sqrt{69}$$

(r has to be positive)

Apr 15, 2022
edited by qjin27  Apr 15, 2022

#1
+66
-1

Since the center is at (-5, 2), The equation of the circle is $$(x+5)^2+(y-2)^2=r^2$$

Then you can multiply out $$x^2+10x+25+y^2-4y+4=r^2$$

$$x^2+y^2+10x-4y+29=r^2 \\ 2x^2+2y^2+20x-8y=2r^2-58 \\ 2r^2-58=80 \\ r^2-29=40 \\ r^2=69 \\ r=\sqrt{69}$$

(r has to be positive)

qjin27 Apr 15, 2022
edited by qjin27  Apr 15, 2022
#2
+66
-3

I'm not sure that this is correct so can anyone double check?

qjin27  Apr 15, 2022
#3
+124696
0

Looks good, qjin  !!!

Apr 15, 2022
edited by CPhill  Apr 15, 2022