Question

Please read below. Thank you.

Answer 1

`\sf\longrightarrow \boxed{\sf x^2+y^2-4x-6y-12=0}`

Explanation:

Here we are given the radius of circle as 5cm and the centre of the circle is (2,3) . We need to find the equation of the circle. Here we can yse the Standard equation of circle to find the equation .

Standard equation of circle :-

`\sf\implies \green{ (x - h )^2+(y-k)^2 = r^2 }`

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

Substitute the respective values ,

`\sf\longrightarrow ( x - 2 )^2 + ( y - 3)^2 = 5^2`

Simplify the whole square ,

`\sf\longrightarrow x^2 + 4 -4x + y^2+9-6y = 25`

Rearrange and add the constants ,

`\sf\longrightarrow x^2 + y^2 -4x -6y +13 = 25`

Subtract 25 on both sides ,

`\sf\longrightarrow x^2 +y^2-4x-6y+13-25=0`

Simplify ,

`\sf\longrightarrow \boxed{\blue{\sf x^2+y^2-4x-6y-12=0}}`

Answer 2

The equation of the circle is given by:

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

where:

(a,b) is the center of the circle

given that the center of our circle is (2,3) with the radius of 5, the equation will be:

(x+2)^2+(y+3)^2=5^2

expanding the above we get:

x^2+4x+4+y^2+6y+9=25

this can be simplified to be:

x^2+4x+y^2+6y=25-13

x^2+y^2+4x+6y=12