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A circle passes through the point (12,0), and is tangent to the y-axis at the point (0,3), as shown. Find the radius of the circle.

 


 

Thanks!

~Noori

 Sep 18, 2020
 #1
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Let center ( r,3) radius:r

(x-r)^2+(y-3)^2=r^2
plug in (12,0)

(12-r)^2+(-3)^2=r^2
r=45/8

 Sep 18, 2020
 #4
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Ok apart from the final step.  Should end up with r = 51/8

Alan  Sep 19, 2020
 #5
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Ooh, thank you! That was very helpful. I thought the steps were right, so thank you!

Noori  Sep 20, 2020
 #2
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 A(0, 3)      B(12, 0)       origin O       center of a circle  C   

 

1/     Line segment  AB = sqrt(AO2 + BO2)

 

 2/    Angle (ABO) = arctan(AO / BO)              ∠ABO = ∠BAC

 

3/     Radius  AC = (AB/2) / cos∠BAC

 Sep 18, 2020
 #3
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Sorry, I don't know trigonometry yet, is there a geometric solution?

Noori  Sep 19, 2020

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