SOLUTION: find an equation of a circle passing through the points A(1,2) and B(1,-2) and touching the line x+2y+5=0?

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Question 967614: find an equation of a circle passing through the points A(1,2) and B(1,-2) and touching the line x+2y+5=0?
Found 2 solutions by josgarithmetic, askhari139:
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
Start this way:

Understand through intuition, the center of the circle is on the x-axis. and this is some unknown point (x,0).

Distance from (x,0) Center to A(1,2) is equal to distance from (x,0) center to B(1,-2). Setup this using the distance formula and find the x value for the center point of the circle.

Now, can you think a way to use the given line touching the circle?
Equivalent line is+y=-x%2F2-5%2F2 and is some general point (x, -(x/2+5/2) );

Unfinished; just a description to begin solving.

Answer by askhari139(3) About Me  (Show Source):
You can put this solution on YOUR website!
To write the equation of any circle, you need a center and a radius. So, if (a,b) is the center and r is the radius of the circle, the circle equation would be
%28x-a%29%5E2+%2B+%28y-b%29%5E2+=+r%5E2
In this problem, you have 2 pieces of information. One, 2 points one the circle and two, a tangent. Let the coordinates of the center be O(a,b) and the radius be r.
Since point A and point B are points on the circle, distance between O and A will be equal to O and B which in turn would be equal to r.
OA = OB = r
=> sqrt%28%28a-1%29%5E2%2B%28b-2%29%5E2%29+=+r....... (1)
sqrt%28%28a-1%29%5E2%2B%28b-%28-2%29%29%5E2%29+=+r.... (2)
Also, since x+2y+5 = 0 touches the circle, it is a tangent to the circle. and hence, distance between centre O and the line will also be equal to r.
=> sqrt%28%28a%2B2b%2B5%29%2Fsqrt%281%5E2%2B2%5E2%29%29+=+r......... (3)
For the 3 variables a,b,r, you have 3 equations. Solve them and you will get the answer.
message me if you have any further problems.