SOLUTION: A circle has equation x^2+y^2-2x+4y-15=0 Find the values for which the line mx+2y-7=0 is a tangent to this circle

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Question 966011: A circle has equation x^2+y^2-2x+4y-15=0
Find the values for which the line mx+2y-7=0 is a tangent to this circle
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
A circle has equation x^2+y^2-2x+4y-15=0
Find the values for which the line mx+2y-7=0 is a tangent to this circle
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x^2+y^2-2x+4y-15=0
(x-1)^2 + (y+2)^2 = 20 is the circle.
mx+2y-7=0
y = mx/2 + 3.5
A point on the tangent lines (2 of them) is A(0,3.5)
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The center of the circle is C(1,-2)
The distance AC =
The tangent point is P
CP = r = sqrt(20)
ACP is a right triangle

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Draw a circle centered at A with r^2 = 11.25
--> x^2 + (y-3.5)^2 = 11.25
The 2 points of intersection of the 2 circles are the tangent points.
x^2 + (y-3.5)^2 = 11.25
x^2 + y^2 - 7y + 12.25 = 11.25
x^2 + y^2 - 7y = -1
x^2 + y^2 - 2x + 4y = 15
------------------------------------------ Subtract
2x - 11y = -16 (the eqn of the line thru the 2 tangent points)
Find the intersection of the line and either circle.
2x - 11y = -16
y = (2x + 16)/11
(x-1)^2 + (y+2)^2 = 20
Sub for y
(x-1)^2 + ((2x+38)/11)^2 = 20
121x^2 - 242x + 121 + 4x^2 + 152x + 1444 = 2420
125x^2 - 90x - 855 = 0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation (in our case ) has the following solutons:



For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=435600 is greater than zero. That means that there are two solutions: .




Quadratic expression can be factored:

Again, the answer is: 3, -2.28. Here's your graph:

==============
x = -2.28 --> y = 1.04 --> (-2.28,1.04)
x = 3 --> y = 2 --> (3,2)
Can you do the rest?
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