I need the answer to this algebra question. 4 x 38/12 + 2
1 answer:
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Answer:
2) c) (x-3)² + (y+2)² = 25
5) x^2 +y^2 -8x -16y +54 = 0
6) x^2 +y^2 -10x -12y +36 = 0
Step-by-step explanation:
2) The standard form equation for a circle is ...
(x -h)^2 +(y -k)^2 = r^2
You are given the center: (h, k) = (3, -2) and a point on the circle. So, the equation will be ...
(x -3)^2 +(y +2)^2 = r^2
Since we know a point on the circle we know that ...
(7 -3)^2 +(1 +2)^2 = r^2 = 16 +9 = 25
So, the circle's equation is ...
(x -3)^2 +(y +2)^2 = 25 . . . . . matches choice C
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5) As in the previous problem, the standard form equation is ...
(x -4)^2 +(y -8)^2 = (-1-4)^2 +(7-8)^2 = 25+1 = 26
To put this in general form, we need to subtract 26 and eliminate parentheses.
x^2 -8x +16 +y^2 -16y +64 -26 = 0
x^2 +y^2 -8x -16y +54 = 0
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6) A circle tangent to the y-axis will have a radius equal to the x-value of the center point.
(x -5)^2 +(y -6)^2 = 5^2
x^2 -10x +25 +y^2 -12y +36 = 25
x^2 +y^2 -10x -12y +36 = 0
