Question

Determine the equation of the circle graphed below.
10
8
-10 -8
8
-6
-2
F
4
2
N
T
-6.
2
10

Answer 1

Answer:

$= > {x}^{2} + {y}^{2} + 6x - 14y + 49 = 0$

Step-by-step explanation:

From the graph we can see that

Diameter = 6 units

=> Radius = 3 units and centre is at (-3,7)

=> Equation is Circle with centre (-3,7) and radius of 3 units will be

$= > {(x + 3)}^{2} + {(y - 7)}^{2} = {3}^{2}$

$= > ( {x}^{2} + 6x + 9) + ( {y}^{2} - 14y + 49) = 9$

$= > {x}^{2} + {y}^{2} + 6x - 14y + 49 = 0$