Question 1026673: Find the equation of the circle which passes through the origin and touches the straight line 4x-3y=5 at (2,1)
Answer by mathmate(429)   (Show Source):
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Question:
Find the equation of the circle which passes through the origin and touches the straight line 4x-3y=5 at (2,1)
Solution:
1. Point of tangency is at (2,1) on 4x-3y=5 means that the centre is located on the perpendicular
3(x-2)+4(y-1)=0
or
3x+4y=10.....(1)
2. Circle passes through T(2,1) and O(0,0) means that segment TO is a chord of the circle. The centre passes lies on its perpendicular bisector.
Mid-point: M((2+0)/2,(1+0)/2)=(1,0.5)
Slope of segment TO: (1-0)/(2-0)=1/2
Slope of perpendicular bisector: -1/(1/2)=-2
Equation of perpendicular bisector: y-0.5=-2(x-1), or
y=-2x+5/2.......(2)
3. Since the centre lies on both equations (1) and (2), it is located by solving the system of equations (1) and (2).
Using substitution of (2) into (1),
3x+4(-2x+5/2)=10 =>
5x=0
x=0, y=5/2
4. Equation of the circle
From the centre located at (0,5/2), and the circle passes through O(0,0)
the circle has the equation:
(x-0)^2+(y-5/2)^2=0^2+(5/2)^2
=>
x^2+(y-5/2)^2=25/4
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