Question

Help please, I need it

Answer 1

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