Question

Determine the equation of the circle graphed below.

Answer 1

• 
  `(x-3)^2+(y+4)^2=29`

Explanation:

To find:-

• The equation of the graphed circle.

Answer:-

We can see that the centre of the given graphed circle is (3,-4) and one of the point on the circumference of the circle is (8,-2) .

Now we can calculate the radius of the circle using these two points as radius is the distance between centre and any point on the circle.

Distance formula:-

`\longrightarrow \boxed{ d =√( (x_2-x_1)^2+(y_2-y_1)^2)} \\`

On substituting the respective values, we have;

`\longrightarrow d = \sqrt{ (8-3)^2+\{ -4-(-2)\}^2}\\`

`\longrightarrow d =√( 5^2 + (-4 +2)^2)\\`

`\longrightarrow d =√( 5^2+2^2) \\`

`\longrightarrow d =√(25+4) \\`

`\longrightarrow d =√( 29)\rm{units}\\`

Now we can use the standard equation of circle to find out the equation of the circle as ,

Standard equation of circle:-

`\longrightarrow \boxed{ (x-h)^2+(y-k)^2=r^2} \\`

where ,

• (h,k) is the centre.
• r is the radius.

On substituting the respective values, we have;

`\longrightarrow (x-3)^2+\{ y-(-4)\}^2 = (√(29))^2 \\`

`\longrightarrow \underline{\underline{\red{(x-3)^3+(y+4)^2=29}}}\\`

This is the required equation of the circle.

Answer 2

(x - 3)² + (y + 4)² = 29

Explanation:

the equation of a circle in standard form is

(x - h )² + (y - k)² = r²

where (h, k ) are the coordinates of the centre and r is the radius

we are given the centre and require to find the radius r

the distance from the centre to a point on the circle gives r

using the distance formula to find r

r =
`\sqrt{(x_(2)-x_{1 )^2+(y_(2)-y_(1))^2 }`

with (x₁, y₁ ) = (3, - 4 ) and (x₂, y₂ ) = (8, - 2 )

r =
`√((8-3)^2+(-2-(-4))^2)`

=
`√(5^2+(-2+4)^2)`

=
`√(25+2^2)`

=
`√(25+4)`

=
`√(29)`

then equation of circle with centre (3, - 4 ) and r =
`√(29)` is

(x - 3)² + (y - (- 4) )² = (
`√(29)` )² , that is

(x - 3)² + (y + 4)² = 29