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Pre-Calc help:

Write the general equation for the circle that passes through the points (1, 1), (1, 3), and (9, 2).
You must include the appropriate sign (+ or -) in your answer. Do not use spaces in your answer.
8x² + 8y²___x___y = 0

1 Answer

3 votes

Answer:


8x^2+8y^2-79x-32y+95=0

Explanation:

Let the equation be :
x^2+y^2+2gx+2fy+c=0

The point (1,1) must satisfy this equation.


1^2+1^2+2g*1+2f*1+c=0\\2g+2f+c=-2---(1)

Similarly, the point (1,3) must satisfy:


1^2+3^2+2g*1+2f*3+c=0\\1+9+2g+6f+c=0\\2g+6f+c=-10---(2)

Also for the point (9,2), we have:


9^2+2^2+2g*9+2f*2+c=0\\81+4+18g+4f+c=0\\18g+4f+c=-85---(3)

Solving the equations (1), (2) and (3) simultaneously, we get:


g=-(-79)/(16),f=-2,c=(95)/(8)

We Substitute this values to get:


x^2+y^2+2((-79)/(16))x+2(-2)y+(95)/(8)=0


x^2+y^2-((79)/(8))x-4y+(95)/(8)=0

Multiply through by 8 to get:


8x^2+8y^2-79x-32y+95=0

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