General Equation of a Circle and Standard form.

distance formula

standard form (x-h)^2 + (y-k)^2=r^2

General form x^2+y^2+ax+by+c=0

Find Center(h,k) find radius of each circle graph its cirlces

21. 2(x-3)^2 +2y^2=8

divide all by 2

(x-3)^2 +y^2=4

Center=(3,0)

r=2

xintercepts {5,1}

31. 2x^2 +2y^2 minus 12x +8y minus 24=0

all i know is minus 24, complete the square, factor

Find standard equation

37. Center (2,3) tangent to x axis

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

How did they get 9?

How does tangent to x axos give me radius of 9?

41. center (-1,3) tangent to line y=2

(x+1)^2 + (y-3)^2 =2

How they get radius of 2?

Re: General Equation of a Circle and Standard form.

Quote:

Originally Posted by

**Melcarthus** 21. 2(x-3)^2 +2y^2=8

divide all by 2

(x-3)^2 +y^2=4

Center=(3,0)

r=2

xintercepts {5,1}

Right.

Quote:

Originally Posted by

**Melcarthus** 31. 2x^2 +2y^2 minus 12x +8y minus 24=0

all i know is minus 24, complete the square, factor

I would start by dividing thorugh by 2. Then complete the squares:

$\displaystyle x^2 - 6x + 9 +y^2 +4y + 4 = 12 + 9 + 4$

$\displaystyle (x-3)^2 + (y+2)^2 = 5^2$

Quote:

Originally Posted by

**Melcarthus** 37. Center (2,3) tangent to x axis

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

How did they get 9?

How does tangent to x axos give me radius of 9?

Given that the circle is tangent to the x axis, the distance from the center to the x-axis is equal to the radius of the circle. The y-coordinate of the center is 3, so that's the radius. Remember that the right jand side is the square of the radius, and 3^2 = 9.

Quote:

Originally Posted by

**Melcarthus** 41. center (-1,3) tangent to line y=2

(x+1)^2 + (y-3)^2 =2

How they get radius of 2?

I think this is in error. The distance from the center at (-1,3) to the line y=2 is 1, so the right hand side should be 1, not 2.