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Help please, I need it

Help please, I need it-example-1
Help please, I need it-example-1
Help please, I need it-example-2
User Kylos
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1 Answer

3 votes

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

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

__

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

__

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

Help please, I need it-example-1
User MikeOnline
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