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.
