Question

How do you write an equation of a circle with Center: (6, 7) Point on Circle: (13 4)?

Answer 1

`x^2+y^2-12x-14y+27=0`

Explanation:

Let say `x and y` is another point on the circle.

`(x-6)^2+(y-7)^2=(13-6)^2+(4-7)^2`
`x^2-12x+36+y^2-14y+49=(7)^2+(-3)^2`
`x^2-12x+36+y^2-14y+49=49+9`
`x^2+y^2-12x-14y+36+49-49-9=0`
`x^2+y^2-12x-14y+27=0`

Answer 2

`x^2+y^2-12x-14y+27=0.`

Explanation:

The Centre `C` of the Circle is `C(6,7)`

The Point `P(13,4)` is on the Circle. Therefore, the Radius `r` of the

Circle is distance `CP`. Using, the Distance Formula,

`r^2=CP^2=(13-6)^2+(4-7)^2=49+9=58.`

We know that the eqn. of a Circle having Centre at #(h,k) and Radius

`r" "is : (x-h)^2+(y-k)^2=r^2`

Therefore, the eqn. of the circle is

`(x-6)^2+(y-7)^2=58, i.e.,`

`x^2+y^2-12x-14y+27=0`

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