SOLUTION: Pls u guys should help me out. The equation of a circle is given as x2 +y2-6x+8y-24=0. Find the equation of the diameter of the circle which passes through the point (-1,1)

Question 984216: Pls u guys should help me out. The equation of a circle is given as x2 +y2-6x+8y-24=0. Find the equation of the diameter of the circle which passes through the point (-1,1)
Found 2 solutions by macston, josgarithmetic:
Answer by macston(5194) (Show Source): You can put this solution on YOUR website!
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Complete the squares

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Standard form for circle:

Where (h,k) is center of circle and r=radius.
So the center of our circle is at (3,-4) with radius=7.
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The equation of the diameter will be the equation through the center and the given point:
Find the line through (-1,1) and (3,-4)
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Find the slope:
m=(change in y)/(change in x)=(1-(-4))/(-1-3)=(5/-4)=-(5/4)
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In slope intercept form:
y=mx+b
y=-(5/4)x+b Put in values of the given point and solve for b.
1=-(5/4)(-1)+b
4/4=5/4+b
-1/4=b
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Write the final equation:
y=-(5/4)x-1/4

Answer by josgarithmetic(39631) (Show Source): You can put this solution on YOUR website!
Not a circle: x2 +y2-6x+8y-24=0

This, a circle: x^2+y^2-6x+8y-24=0

Complete the Square to put into standard form: .

The diameter must include the center of the circle, (3,-4), and the given point (-1,1) must be on the LINE that contains a diameter of the circle (as asked).

Same now as, find the equation for the line which passes through (3,-4) and (-1,1). Using point-slope form and only determining the slope of those two points, and choosing (-1,1), equation is .
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