Question

A circle is defined by x^2+y^2+2gx+2fy-15=0. The gradient of the tangent to the circle at the point (3,2) is -1/2. Find the value of g and f​

Answer 1

Answer:

  (g, f) = (-1, 2)

Step-by-step explanation:

The equation can be put into standard form:

  (x^2 +2gx) +(y^2 +2fy) = 15

  (x^2 +2gx +g^2) +(y^2 +2fy +f^2) = 15 +g^2 +f^2

  (x +g)^2 +(y +f)^2 = 15 +g^2 +f^2 . . . . equation in standard form

The point (3, 2) is on the circle, so one of the equations we have is ...

  (3+g)^2 +(2 +f)^2 = 15 +g^2 +f^2

  9 +6g +g^2 +4 +4f +f^2 = 15 +g^2 +f^2

  6g +4f = 2 . . . . . . . subtract g^2 +f^2 +13

  3g +2f = 1 . . . . . . . divide by 2

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A line through the center of the circle is perpendicular to the tangent at the point of tangency. That line will have a slope that is the negative reciprocal of the tangent's slope: m = -1/(-1/2) = 2. Then in point-slope form the equation of the line through the center is ...

  y -2 = 2(x -3)

  2x -y = 4 . . . . rearrange to standard form

The point at the center of the circle, (-g, -f), satisfies this equation, so we have another relation for f and g:

  2(-g) -(-f) = 4

  2g -f = -4 . . . . . . multiply by -1

Adding twice this equation to the one above, we have ...

  2(2g -f) +(3g +2f) = 2(-4) +(1)

  7g = -7 . . . . . . . . . simplify

  g = -1 . . . . . . . . . . divide by 7

  f = 2g +4 = 2(-1) +4 = 2 . . . find the corresponding value of f

The values of f and g are: f = 2 and g = -1.