What is (x+3)(x+4)+(y-4)(y-3)=0 ?

The solution to (x+3)(x+4)+(y-4)(y-3)=0 is Circle with (a,b)=(-7/2 , 7/2),r=sqrt(1/2)

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(x+3)(x+4)+(y-4)(y-3)=0

(x + 3) (x + 4) + (y - 4) (y - 3) = 0

(a, b) = (- \frac{7}{2}, \frac{7}{2}), r = \frac{1}{2}

Solution steps

(x + 3) (x + 4) + (y - 4) (y - 3) = 0

Rewrite

(x + 3) (x + 4) + (y - 4) (y - 3) = 0

in the form of the standard circle equation

(x - (- \frac{7}{2}))^{2} + (y - \frac{7}{2})^{2} = (\frac{1}{2})^{2}

Therefore the circle properties are:

(a, b) = (- \frac{7}{2}, \frac{7}{2}), r = \frac{1}{2}

(x + 5)^{2} + (y + 8)^{2} = 20

4 x^{2} + 4 y^{2} = 4

x^{2} + y^{2} - 8 x - 3 = 0

364 + 28 y + y^{2} + x^{2} = - 26 x

x^{2} + y^{2} - 49 < 0

What is (x+3)(x+4)+(y-4)(y-3)=0 ?
The solution to (x+3)(x+4)+(y-4)(y-3)=0 is Circle with (a,b)=(-7/2 , 7/2),r=sqrt(1/2)