Question 483622: Find the equation of the circle. Touching the x-axis and passing through the points (3,1),(10,8). ty
Answer by Edwin McCravy(20060)   (Show Source):
You can put this solution on YOUR website! Find the equation of the circle. Touching the x-axis and passing through the points (3,1),(10,8).
We plot those points:
We can guess how to draw this circle using a compass. There are
two possible circles:
Let the center of the circle be (h,k), then since the circle touches
the x-axis, we draw a green vertical radius down to the x-axis:
Because it touches the x-axis, we can see from the graph above, k,
which is the center's y-coordinate is the same as the radius r.
The equation of a crcle with center (h,k) and radius r has equation:
(x - h)² + (y - k)² = r²
And since k = r, it is
(x - h)² + (y - r)² = r²
x² - 2hx + h² + y² - 2ry + r² = r²
x² - 2hx + h² + y² - 2ry = 0
We now substitute each of the given points in the equation:
Substituting (x,y) = (3,1)
3² - 2h(3) + h² + 1² - 2r(1) = 0
9 - 6h + h² + 1 - 2r = 0
10 - 6h + h² - 2r = 0
Substituting (x,y) = (10,8)
x² - 2hx + h² + y² - 2ry = 0
10² - 2h(10) + h² + 8² - 2r(8) = 0
100 - 20h + h² + 64 - 16r = 0
164 - 20h + h² - 16r = 0
So we have this system of equations:
10 - 6h + h² - 2r = 0
164 - 20h + h² - 16r = 0
We eliminate r by multiplying the first equation by -8
and adding the two equations:
-80 + 48h - 8h² + 16r = 0
164 - 20h + h² - 16r = 0
-------------------------
84 + 28h - 7h² - 0
Arrange in descending powers of h
-7h² + 28h + 84 = 0
Divide through by -7
h² - 4h - 12 = 0
Factor:
(h - 6)(h + 2) = 0
h - 6 = 0; h + 2 = 0
h = 6; h = -2
To find the value of r corresponding to h = 6, substitute
in
10 - 6h + h² - 2r = 0
10 - 6(6) + 6² - 2r = 0
10 - 36 + 36 - 2r = 0
10 - 2r = 5
-2r = -10
r = 5
So one possibility is the amaller circle
(x - h)² + (y - r)² = r²
which becomes
(x - 6)² + (y - 5)² = 5²
That's the standard form, which when multiplied out gives
the general form:
x² + y² - 12x - 10y + 36 = 0
---
To find the value of r corresponding to h = -2, substitute
h = 2 in
10 - 6h + h² - 2r = 0
10 - 6(-2) + (-2)² - 2r = 0
10 + 12 + 4 - 2r = 0
26 - 2r =
-2r = -26
r = 13
And the other possibility is the larger circle which is
(x - h)² + (y - r)² = r²
(x + 2)² + (y - 13)² = 13²
That's the standard form, which when multiplied out gives
the general form:
x² + y² + 4x - 25y + 4 = 0
Two possible solutions.
Edwin
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