# What is the radius of a circle of the function x^2+y^2+2x-6y=10  at the center (-1,3).

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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The equation of the circle which has the center in the point C (-1,3), is written as following:

[x-(-1)]^2 + (y-3)^2 = R^2, where R is the radius of the circle.

We'll develop the binomials and we'll have:

x^2+2x+1+y^2-6y+9-R^2=0

The expression given in enunciation is:

x^2+y^2+2x-6y-10=0

The 2 expressions have to describe the same equation of a circle, so they have to be identically:

x^2+2x+1+y^2-6y+9-R^2=x^2+y^2+2x-6y-10

After reducing similar terms, we'll get:

10-R^2=-10

R^2=20

R=sqrt20

R=2sqrt5

neela | High School Teacher | (Level 3) Valedictorian

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Writing the given equation x^2+y^2+2x-6y =10 like :-

(x-h)^2+(y-k)^2 = r^2, which is the equation of a circle with  centre (h,k) and radius r , we get the centre and radius.

Now let us try to rewrite the equation x^2+2x +y^2-6y = 10.

To make x^2+2x a square  we add 1^2 and subtract -1^2. Similarly to make y^2-6y a  square, we add 3^2 and then subtract 3^2. So,the given equation becomes:

(x^2+2x+1^2)-1^2 + (y^2-6y+3^2) -3^2 = 10. Or

(x+1)^2 +(y-3)^2 = 10+1^2+3^2 . Or

(x- -1)^2 +(y-3)^2 = 20 = (sqrt20)^2. Identifying this equation with (x-h)^2+(y-k)^2 = r^2, we get:

Therefore (h, k) = (-1, 3) is the centre of the circle and r = sqrt20 = 2sqrt5 = 4.472136 approximately.

krishna-agrawala | College Teacher | (Level 3) Valedictorian

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In a circle with the equation of the form:

x^2 + y^2 - 2ax - 2by + c = 0

The center of the circle is given by (a, b).

and radius of the circle is given by:

Radius = (c - a^2 - b^2)^(1/2)

The given equation is:

x^2 + y^2 + 2x - 6y + 10 = 0

Therefore:

a = -1, b = 3, and c = 10

This center of the circle is (-1, 3).

As we can see this is same as the center given in the question.

Radius = (c - a^2 - b^2)^(1/2)

= [10 + (-1)^2 + 3^2]^1/2

= (10 + 1 + 9)^1/2 = 20^1/2 = 4.4721

Please note that we could have calculated the length of the radius even if the center of the circle was not given.