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This is a continuation of tutorial on equations of circles.
Example 5: Find the equation of the circle such that the three points A(0 , 4), B(3 , 5) and D(7 , 3) are on the circle.
Solution to Example 5:
- The distance from the center C(h , k) of the circle to each of the points A, B and D is constant and equal to the radius r of the circle. Write three equations stating that these distances are equal to the radius r.
d(A,C) = sqrt[(h - 0)2 + (k - 4)2] = r
d(B,C) = sqrt[(h - 3)2 + (k - 5)2] = r
d(D,C) = sqrt[(h - 7)2 + (k - 3)2] = r
- Write that d(A,c) = d(B,C) and d(A,C) = d(D,C).
sqrt[(h - 0)2 + (k - 4)2] = sqrt[(h - 3)2 + (k - 5)2]
sqrt[(h - 0)2 + (k - 4)2] = sqrt[(h - 7)2 + (k - 3)2]
- Square each side of each equation.
(h - 0)2 + (k - 4)2 = (h - 3)2 + (k - 5)2
(h - 0)2 + (k - 4)2 = (h - 7)2 + (k - 3)2
- Expand the squares in the above equations and simplify.
-8k + 16 = -6h + 9 -10k + 25
-8k + 16 = -14h + 49 -6k +9
- Write the above system of equations in standard form.
2k + 6h = 18
-2k + 14h = 42
- Use the method of addition to solve the system.
20h = 60
h = 3
- Substitute h by its value 6 in one of the equations to obtain k.
k = 0
- We now use one of the distance formula in part a above to find the radius r.
r = sqrt[(3 - 0)2 + (0 - 4)2]
= 5
- The equation of the circle is given by.
(x - h)2 + (y - k)2 = r2
(x - 3)2 + y2 = 25
Shown below is the graph of the circle with the three points.
Matched Exercise: Find the equation of the circle such that the three points A(-5 , 0), B(1 , 0) and D(-2 , -3) are on the circle.
Answer.
More links and references related to the above topics.
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