Question 1028984: Write the general and standard form of the equation of the circle satisfying the given conditions.
With center (2, -1) and radius 3
With center (-4, 5) and passing through (2, 1)
Having the points (-1, -1) and ( 3, 4 ) as ends of the diameter
Center at ( 4, 1 ) and touching the y- axis
With center at (-2, -1 ) and tangent to the line 4x-3y = 12
Passing through ( 1,2), (2,3) and (-2, 1)
Passing through (4, 6), (-2,-2) and (-4, 2)
Tangent to the line 3x- 4y - 5 = 0 at (3,1) and passing through ( -3, -1)
Inscribed in a triangle with sides on the lines x – 3y = -5, 3x + y = 1 and 3x – y = -11.
Find the general equation of the tangent to the circle
(x - 〖3)〗^2 + (y - 〖5)〗^2 = 64 at point ( 3, 3 )
x^2 + y^2 – 2x – 24 = 0 at point ( -2, -4 )
x^2 + y^2 – 2x – 24 = 0 at point ( -2, -4 )
x^2 + y^2 – 6x – 4y - 28 = 0 parallel to the line 4x + 5y = 8
x^2 + y^2 – 4y – 9 = 0 through ( 5 , -1 )
x^2 + y^2 – 6x – 2y + 5 = 0 through ( -3, 6 )
For each pair of equations of circles, find the general equation of the radical axis
x^2 + y^2 – 14y + 40 = 0 and x^2 + y^2 = 4
x^2 + y^2 – 14x – 12y + 65 = 0 and x^2 + y^2 – 6x – 4y + 3 = 0
x^2 + y^2 – 12x + 14y + 60 = 0 and x^2 + y^2 + 6x + 4y - 3 = 0
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! It's not likely anyone will do all these. It might take a day or 2, but it will scroll off the 1st page and be gone.
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Write the general and standard form of the equation of the circle satisfying the given conditions.
With center (2, -1) and radius 3
A circle or radius r with its center at (h,k) is (x-h)^2 + (y-k)^2 = r^2
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With center (-4, 5) and passing through (2, 1)
Having the points (-1, -1) and ( 3, 4 ) as ends of the diameter
Find the midpoint, that's the center. r = 1/2 the distance between the 2 points. Then it's the same as the 1st problem.
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Center at ( 4, 1 ) and touching the y- axis
It's 4 units from the y-axis --> r = 4
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With center at (-2, -1 ) and tangent to the line 4x-3y = 12
Passing through ( 1,2), (2,3) and (-2, 1)
Passing through (4, 6), (-2,-2) and (-4, 2)
Tangent to the line 3x- 4y - 5 = 0 at (3,1) and passing through ( -3, -1)
Inscribed in a triangle with sides on the lines x – 3y = -5, 3x + y = 1 and 3x – y = -11.
Find the general equation of the tangent to the circle
(x - 〖3)〗^2 + (y - 〖5)〗^2 = 64 at point ( 3, 3 )
x^2 + y^2 – 2x – 24 = 0 at point ( -2, -4 )
x^2 + y^2 – 2x – 24 = 0 at point ( -2, -4 )
x^2 + y^2 – 6x – 4y - 28 = 0 parallel to the line 4x + 5y = 8
x^2 + y^2 – 4y – 9 = 0 through ( 5 , -1 )
x^2 + y^2 – 6x – 2y + 5 = 0 through ( -3, 6 )
For each pair of equations of circles, find the general equation of the radical axis
x^2 + y^2 – 14y + 40 = 0 and x^2 + y^2 = 4
x^2 + y^2 – 14x – 12y + 65 = 0 and x^2 + y^2 – 6x – 4y + 3 = 0
x^2 + y^2 – 12x + 14y + 60 = 0 and x^2 + y^2 + 6x + 4y - 3 = 0
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PS Don't put spaces in the (x,y) points.
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