x^2 + y^2 = r^2
2x + 2y dy/dx = 0 (r is a constant)
2y dy/dx = -2x
dy/dx = -x/y
so if a line is tangent to a circle, the slope of the line will be equal to the slope of the circle at the point of tangency. So if the circle is tangent to a line with a slope of 1/2 then -x/y = 1/2 this gives you the equation -2x = y, so you can substitute for y in the equation x^2 + y^2 = r^2.
Note, a circle reaches a given slope at exactly two points, so you must check both points to see which falls into the equation of the tangent line. (basicly subsitute the x and y values into the linear equation and see if it comes out true).
]]>Anyway, the formula for a circle centred at the origin is x² + y² = r², where r is the radius.
To move the centre, we need to change what we square accordingly.
We need to add or subtract how much the centre would need to move in each direction to get to the origin.
Your circle is centred at (-3, 3), so to be centred at the origin, it would need to move 3 in the x direction and -3 in the y direction.
And your radius is 3, so you want the right hand side to be 9.
Therefore, the equation for your circle is (x+3)² + (y-3)² = 9.
]]>sorry my english is not that much so i'll insert a picture :
sorry again for this bad english !
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