Author Topic: Circles (Read 636 times) Tweet Share

Shubhashish Banerjee · « **on:** January 15, 2013, 11:12:32 am »

Hey guys how do you find the equation of a circle which passes through the points (3,1), (8,2), and (2,6) using the general equation of a circle. Thanks

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polar · « **Reply #1 on:** January 15, 2013, 11:20:11 am »

$\begin{aligned} (3-h)^2 + (1-k)^2 &= R \\ (8-h)^2 + (2-k)^2 &= R \\ (2-h)^2 + (6-k)^2 &= R \end{aligned}$

Solving for h,k and R:

Spoiler

$\begin{aligned} \text{Equate equations 1 and 2} \\ (3-h)^2 + (1-k)^2 & = (8-h)^2 + (2-k)^2 \\ 10h + 2k & =58 \\ k & = -5h + 29 \\ \text{Equate equations 1 and 3} \\ (2-h)^2 + (6-k)^2 & = (3-h)^2 + (1-k)^2 \\ 2h-10k & = -30 \\ \text{Substitute\;} k = -5h+29 \\ 2h-10(-5h+29) & = -30 \\ h & = 5 \\ k & = -5(5) + 29 \\ k & = 4 \\ \text{Substitute \;} h=5, k=4 \; \text{into equation 1 to find\;} R \\ (3-5)^2 + (1-4)^2 & = (2)^2 + (3)^2 \\ & =13 \\ \text{Hence, the equation of the circle is \;} (x-5)^2 + (y-4)^2 &= 13\end{aligned}$

DJing · « **Reply #2 on:** January 15, 2013, 11:21:50 am »

Easiest method is to solve on the calculator using the general formula for circles and 3 variables

I.e. replace x&y with coords

You should then get h, k & z

Shubhashish Banerjee · « **Reply #3 on:** January 15, 2013, 11:35:43 am »

Awesome! Thanks @Polar and @Djing. That helps a lot.

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Author Topic: Circles (Read 636 times) Tweet Share

Shubhashish Banerjee

Circles

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