SOLUTION: I have been struggling on this math problem and I was wondering if someone could help me? Please and Thank You!! I would deeply appreciate!! Circles Find an equation of the cir

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Question 190519This question is from textbook Algebra and Trigonometry Structure and Method book 2
: I have been struggling on this math problem and I was wondering if someone could help me? Please and Thank You!! I would deeply appreciate!!
Circles
Find an equation of the circle described.
Center in quadrant four; tangent to the lines x=1, x=9 and y=0 This question is from textbook Algebra and Trigonometry Structure and Method book 2

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Just a quick reminder: the equation of any circle is in the form %28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2 where (h,k) is the center and "r" is the radius.


If you draw the lines x=1 and x=9, you'll get two vertical lines at x=1 and x=9. Since the circle is tangent to both these lines, and the two lines are parallel, this means that endpoints of the diameter touches these lines. So the endpoints of the diameter are the points (1,y) and (9,y) where "y" is some unknown number


So this basically means that the endpoints of the diameter have x-coordinates of 1 and 9. Now let's find the midpoint of the diameter to get the center of the circle:


First, add 1 to 9 to get 10. Now take half of 10 to get 5.


So the midpoint of the diameter has an x-coordinate of 5. This means that the x-coordinate of the center is x=5 which also means that h=5



Take note that the distance from 5 to 9 is 4 and the distance from 1 to 5 is 4. So this means that the radius of the circle is 4 units. So we've found that r=4


Now because the circle is also tangent to y=0, this means that the center is also 4 units away from points of the form (x,0). The only two points that have x-coordinates of 5 and are 4 units away from y=0 are (5,-4) and (5,4). But we have to remember that the problem states that: "Center in quadrant four". So this eliminates the second point leaving the only possible center of (5,-4)


So the center of the circle is (5,-4) and the radius is 4 units. This means that h=5, k=-4 and r=4


%28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2 Start with the given formula


%28x-5%29%5E2%2B%28y-%28-4%29%29%5E2=4%5E2 Plug in h=5, k=-4 and r=4


%28x-5%29%5E2%2B%28y%2B4%29%5E2=4%5E2 Simplify


%28x-5%29%5E2%2B%28y%2B4%29%5E2=16 Square 4 to get 16



So the equation of the circle with the given properties is %28x-5%29%5E2%2B%28y%2B4%29%5E2=16


Here's a drawing to confirm



Graph of %28x-5%29%5E2%2B%28y%2B4%29%5E2=16 (with center (5,-4)) with the blue tangent lines x=1, x=9 (vertical lines) and y=0 (blue line overlapping the x-axis)