Question

NO LINKS!!!! Find the equation of the circle below​

Answer 1

Answer:

$(x+3)^2+(y-2)^2=9$

Step-by-step explanation:

Equation of a circle

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

where:

• (a, b) is the center
• r is the radius

From inspection of the diagram, the center of the circle appears to be at point (-3, 2), although this is not very clear.  Therefore, a = -3  and  b = 2.

Substitute these values into the general form of the equation of a circle:

$\implies (x-(-3))^2+(y-2)^2=r^2$

$\implies (x+3)^2+(y-2)^2=r^2$

Again, from inspection of the diagram, the maximum vertical point of the circle appears to be at y = 5.  Therefore, to calculate the radius, subtract the y-value of the center point from the y-value of the maximum vertical point:

⇒ radius (r) = 5 - 2 = 3

Substitute the found value of r into the equation:

$\implies (x+3)^2+(y-2)^2=3^2$

Therefore, the final equation of the given circle is:

$\implies (x+3)^2+(y-2)^2=9$

Answer 2

Centre(-3,2)

diameter=6 units [0-(-6)=6units ]

radius=6/2=3

Equation of circle

$\\ \rm\Rrightarrow (x-h)^2+(y-k)^2=r^2$

• For centre(h,k) and radius r

So our equation

$\\ \rm\Rrightarrow (x+3)^2+(y-2)^2=3^2$

$\\ \rm\Rrightarrow (x+3)^2+(y-2)^2=9$