What is (x-1)(4-x)+(y+2)(6-y)=0 ?

The solution to (x-1)(4-x)+(y+2)(6-y)=0 is Circle with (a,b)=(5/2 ,2),r=(sqrt(73))/2

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(x-1)(4-x)+(y+2)(6-y)=0

(x - 1) (4 - x) + (y + 2) (6 - y) = 0

(a, b) = (\frac{5}{2}, 2), r = \frac{73}{2}

Solution steps

(x - 1) (4 - x) + (y + 2) (6 - y) = 0

Rewrite

(x - 1) (4 - x) + (y + 2) (6 - y) = 0

in the form of the standard circle equation

(x - \frac{5}{2})^{2} + (y - 2)^{2} = (\frac{73}{2})^{2}

Therefore the circle properties are:

(a, b) = (\frac{5}{2}, 2), r = \frac{73}{2}

x^{2} + 8 x + y^{2} + 4 y - 4 = 0

0 = 9 - x^{2} - y^{2}

5 x^{(2)} + 5 y^{(2)} + 10 x + 2 y + 3 = 0

x^{2} + y^{2} = 16 x

(x + 8)^{2} + (y + 3)^{2} = 4

What is (x-1)(4-x)+(y+2)(6-y)=0 ?
The solution to (x-1)(4-x)+(y+2)(6-y)=0 is Circle with (a,b)=(5/2 ,2),r=(sqrt(73))/2