Can anyone help me with this problem??

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  (a)  4/3

  (b)  y -3 = 4/3(x -1)

  (c)  y -3 = -3/4(x -1)

  (d)  r = 5

Step-by-step explanation:

a) The slope is given by the slope formula:

  m = (y2 -y1)/(x2 -x1)

  m = (7 -3)/(4 -1) = 4/3

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b) The radius is normal to the circle. The point-slope form of the equation for a line can be useful here:

  y -k = m(x -h) . . . . . line with slope m through point (h, k)

For slope 4/3, the line through point (1, 3) will have the equation ...

  y -3 = 4/3(x -1) . . . . point-slope equation of the normal

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c) The tangent is perpendicular to the radius. It will have a slope that is the opposite reciprocal of the slope of the radius: -1/(4/3) = -3/4.

  y -3 = -3/4(x -1) . . . . point-slope equation of the tangent

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d) The radius can be found from the distance formula.

  d = √((x2 -x1)² +(y2 -y1)²)

  d = √((4 -1)² +(7 -3)²) = √(3² +4²) = √25 = 5

The radius of the circle is 5.

if x = 1, then the answer is 3.5