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Question 1193536: 1. Find an equation of a circle passing through the points A(1,2) and B(1,-2) touching the line x+2y+5=0.
2. Find the center and radius of 5x^2+5y^2+24+36y+10=0
Found 2 solutions by Alan3354, greenestamps:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! 2. Find the center and radius of 5x^2+5y^2+24+36y+10=0
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Complete the squares for x and for y:
5x^2+5y^2+24+36y+10=0
5x^2 + 5y^2 + 36y = -34
x^2 + y^2 + 7.2y = -6.8
x^2 + y^2 + 7.2y + 12.96 = -6.8 + 12.96 = 6.16
x^2 + (y + 3.6)^2 = 6.16
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Center at (0,-3.6)
Answer by greenestamps(13195)  (Show Source):
You can put this solution on YOUR website!
The other tutor answered your second question, which is rather basic.
Your first question is much more interesting.
The circle passes through A(1,2) and B(1,-2), so AB is a chord of the circle.
The perpendicular bisector of any chord of a circle passes through the center of the circle; the perpendicular bisector of AB is the x-axis. So the center of the circle is (a,0).
(1) The radius of the circle is the distance from (a,0) to (1,2).
(2) The radius of the circle is also the (shortest) distance from (a,0) to the line x+2y+5=0.
The shortest distance from a point (p,q) to a line with equation Ax+By+C=0 is given by the formula
For this problem the distance from (a,0) to the line x+2y+5=0 is
Square both expressions for the radius and set them equal to each other to solve for a.
There appear to be 2 solutions; a=0 or a=5.
If a=0, then
and the equation of the circle is
If a=5, then
and the equation of the circle is
A graph, showing the line x+2y+5=0 and the two circles tangent to it that both pass through the given point (1,2) and (1,-2):
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