Question

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

Answer 1

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`

Answer 2

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

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`