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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(39630)
      (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
  and is some general point  (x, -(x/2+5/2) ); 
 Unfinished; just a description to begin solving.
Answer by askhari139(3)
      (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 
  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
 =>
  ....... (1) 
  .... (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.
 =>
  ......... (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.
 
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