Equation of Circle Passing Through Given Point and Line with Given Radius

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To find the equation of a circle that passes through point A(-3,1) with a radius of 2 and a center on the line 2x-3y+3=0, two initial equations are derived from the point and line conditions. The distance from point A to the center must equal the radius, leading to a third equation involving the center's coordinates. A more efficient approach suggested is to express the center's x-coordinate as a variable, calculate the corresponding y-coordinate, and set the distance to A equal to 2. This method simplifies the problem to a single variable equation, potentially yielding two solutions for the center.
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Homework Statement


Find an equation of the circle passing through:

A(-3,1) with radius 2 and centre on the line 2x-3y+3=0

Homework Equations


x2+y2+2gx+2fy+c=0
r2=g2+f2-c

The Attempt at a Solution


using this equation , i have found 2 equations
-6g+2f+c=-10 by putting (-3,1)

-2g+3f+3=0 (-g,-f) in the linear equation which passes through the centre

how can i use radius information here to get the third equatio n ?
 
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If A(-3,1) is on the circle, what do you know about its distance to the center?
Alternatively: Just plug in the radius in your second "relevant equation".

(It would be easier to start with the line of the centers and just one unknown variable, but your approach works as well).
 
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alijan kk said:
and centre at 2x-3y+3=0
I think you meant "and centre on (the line) 2x - 3y + 3 = 0."
 
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mfb said:
If A(-3,1) is on the circle, what do you know about its distance to the center?
Alternatively: Just plug in the radius in your second "relevant equation".

(It would be easier to start with the line of the centers and just one unknown variable, but your approach works as well).
(-3,1) (-g,-f) distance between them is equal to r

g2+f2+2f+6g+10=r2

g2+f2-c=r2

i get a weird anwer by solving simultaneously ,,, like g=0
 
mfb said:
If A(-3,1) is on the circle, what do you know about its distance to the center?
Alternatively: Just plug in the radius in your second "relevant equation".

(It would be easier to start with the line of the centers and just one unknown variable, but your approach works as well).
(-3,1) (-g,-f) distance between them is equal to r

g2+f2+2f+6g+10=r2

g2+f2-c=r2

i get a weird anwer by solving simultaneously ,,, like g=0

which is the better and easier strategy ?
 
I would start with the suggestion at the end of my post. The center is on the line 2x-3y+3=0? Let its x value be x0, calculate its y-value, calculate the distance to A(-3,1) and require that this distance is 2. A single variable, a single equation to satisfy.
 
@alijan kk: Are you going to share with us what you got? You should have two answers.
 
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