Find the equation of a circle with radius 5 whose centre lies on x axis and passes through the point (2,3)
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Answered by
21
The distance from the x-axis to (2,3) is the radius, r = 5.
There are 2 points in the x-axis that are 5 units from the point (2,3).
The points are (x,0).
d2= 25 = diffy2+diffx2=3 2+(2-x)2
x2-4x+13=25
x2-4x-12=0
(x - 6)*(x + 2) = 0
x = -2, x = 6
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Circle 1 (x+2)2+y2=5 2
Circle 2 (x-6)2+y2=5 2
There are 2 points in the x-axis that are 5 units from the point (2,3).
The points are (x,0).
d2= 25 = diffy2+diffx2=3 2+(2-x)2
x2-4x+13=25
x2-4x-12=0
(x - 6)*(x + 2) = 0
x = -2, x = 6
------------
Circle 1 (x+2)2+y2=5 2
Circle 2 (x-6)2+y2=5 2
Answered by
33
If the center lies on x axis, y coordinate is zero. let the centre be (a, 0).
(x-a)² + (y-0)² = 5²
(2,3) lies on the circle. So, (2-a)²+3² = 5²
(2 - a)² = 25 - 9 = 16
2 - a = +4 or -4 => a = -2 or 6
Equation of the circle :
(x+2)² + y² = 5² with center at (-2, 0)
or (x-6)² + y² = 5² with center at (6,0)
(x-a)² + (y-0)² = 5²
(2,3) lies on the circle. So, (2-a)²+3² = 5²
(2 - a)² = 25 - 9 = 16
2 - a = +4 or -4 => a = -2 or 6
Equation of the circle :
(x+2)² + y² = 5² with center at (-2, 0)
or (x-6)² + y² = 5² with center at (6,0)
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