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The points (−3, 5) and (7, 9) are the ends of the diameter of a circle. a) Find the equation of the circle. b) Find the equation of the normal to the circle at (3, 4).

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2 parts
a

(x−2)2+(y−7)2=29(x - 2)^2 + (y - 7)^2 = 29

b

3x+y−13=03x + y - 13 = 0

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Here are the solutions to the questions.

Part 11

The points (−3,5)(-3, 5) and (7,9)(7, 9) are the ends of the diameter of a circle.

Part a

(x−2)2+(y−7)2=29(x - 2)^2 + (y - 7)^2 = 29

Find the equation of the circle.

  1. Find the coordinates of the center of the circle $(h, k)$.

    The center is the midpoint of the diameter.
    Let (x1,y1)=(−3,5)(x_1, y_1) = (-3, 5) and (x2,y2)=(7,9)(x_2, y_2) = (7, 9).

    h=x1+x22=−3+72=42=2h = \frac{x_1 + x_2}{2} = \frac{-3 + 7}{2} = \frac{4}{2} = 2 k=y1+y22=5+92=142=7k = \frac{y_1 + y_2}{2} = \frac{5 + 9}{2} = \frac{14}{2} = 7

    The center of the circle is (2,7)(2, 7).

  2. Find the radius squared (r2r^2) of the circle.

    The radius is the distance from the center to one of the endpoints of the diameter. We'll use the center (2,7)(2, 7) and the point (−3,5)(-3, 5).

    r2=(x−h)2+(y−k)2r^2 = (x - h)^2 + (y - k)^2 r2=(−3−2)2+(5−7)2r^2 = (-3 - 2)^2 + (5 - 7)^2 r2=(−5)2+(−2)2r^2 = (-5)^2 + (-2)^2 r2=25+4r^2 = 25 + 4 r2=29r^2 = 29

    The radius squared is 29.

  3. Write the equation of the circle using the standard form (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2.

    (x−2)2+(y−7)2=29\boxed{(x - 2)^2 + (y - 7)^2 = 29}

Part b

3x+y−13=03x + y - 13 = 0

Find the equation of the normal to the circle at $(3, 4)$.

  1. Understand that the normal to a circle at any point on its circumference is the line that passes through that point and the center of the circle.

    The point on the circle is $(3, 4)$.
    The center of the circle is $(2, 7)$ (from part a).

  2. Calculate the gradient of the normal (mnm_n) using the two points (3,4)(3, 4) and (2,7)(2, 7).

    mn=y2−y1x2−x1=7−42−3m_n = \frac{y_2 - y_1}{x_2 - x_1} = \frac{7 - 4}{2 - 3} mn=3−1m_n = \frac{3}{-1} mn=−3m_n = -3

  3. Use the point-slope form of a linear equation, y−y1=m(x−x1)y - y_1 = m(x - x_1), with the point (3,4)(3, 4) and the gradient mn=−3m_n = -3.

    y−4=−3(x−3)y - 4 = -3(x - 3)

  4. Simplify and rearrange the equation into the general form Ax+By+C=0Ax + By + C = 0.

    y−4=−3x+9y - 4 = -3x + 9 3x+y−4−9=03x + y - 4 - 9 = 0 3x+y−13=0\boxed{3x + y - 13 = 0}
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