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Write the standard form of the equation of the circle with the given characteristics center: (-4,4); solution?

write the standard form of the equation of the circle with the given characteristics center: (-4,4); solution point: (-2,-6)

This is the competed question.
• 6 months ago

Best Answer - Chosen by Voters

(x+4)^2 + (y-4)^2 = r^2
4+100 = r^2
(x+4)^2+(y-4)^2 = 104
• 6 months ago
100% 1 Vote
• 1. First draw graph with center at (-4,4).

2. Place the point (-2,-6) on the graph.

3. You have a triangle with leg on y axis from -4 going to -2 = -6

4. Then there's the shorter leg of the triangle on x axis from -4 going to -2 . (-4) - (-2) = -2 .

5. Get the hypotenuse which is the radius; square root of [(-6) squared + (-2) ]

square root of [ 36 + 4 ]

square root of 40

equation is (2) times square root of 10 = square root of [x square + y squared ]

Source(s):

analytic geometry
• 6 months ago
• We know that the equation of circle is this type..

(x-a)^2 + (y-b)^2=r
^2,

We use the property that
'distance b/w centre and solution pt(pt at circumference) equals to radius'.

So r^2={-4-(-2)}^2 + {4-(-6)}^2

u will find..

r^2=104.

Now by standard equation,
(x-a)^2 + (y-b)^2=r
^2,

a=-4,b=4,r^2=104.
So Eqn is

(x-4)^2 + (y-4)^2=104,
or
x^2 + y^2 + 8x-8y-72=0
• Edited 6 months ago