SOLUTION: 1. Find the equations of the tangents with y-intercept 5 to the circle x^2 + y^2 = 5. What I know: 1st equation: y= mx + 5 2nd equation: x^2 + y^2 = 5 Please please help me.

Question 971021: 1. Find the equations of the tangents with y-intercept 5 to the circle x^2 + y^2 = 5.
What I know:
1st equation: y= mx + 5
2nd equation: x^2 + y^2 = 5
Please please help me...I don't know how to sub in the points...

Found 3 solutions by josgarithmetic, Boreal, Alan3354:
Answer by josgarithmetic(39620)   (Show Source): You can put this solution on YOUR website!
You would be looking for EXACTLY one point of intersection. If two points of intersection, neither would be a point of tangency.

The rest might be messy, but you want the discriminant to be ZERO. The single variable will be m and you just....(?) solve for m using the discriminant.
Answer by Boreal(15235)   (Show Source): You can put this solution on YOUR website!
Hard to show this without drawing, but here goes.
The line starts at (0,5) and has equal tangents to the circle, one in the first quadrant and one in the second quadrant. Let's look at the first quadrant.
The circle has radius sqrt (5), not 5.
The line drawn tangent to the circle forms a perpendicular line with the radius. That means we have a right triangle, with the hypotenuse the y-axis, and 5 units, the short leg the radius (sqrt 5), and the third leg, the tangent, with a length of 2 sqrt 5 by the Pythagorean theorem.
Using the distance formula.
x^2+y^2=25
We know that the distance from the point on the circle to the end of the line is
sqrt {(x-0)^2 + (y-5)^2} using the distance formula.
But this is equal to sqrt 20.
Therefore, square both sides and x^2 +(y-5)^2= 20
x^2 +y^2 -10y+ 25=20. But x^2 +y^2 =5
5-10y +25 =20
-10y=-10
y=1
x=2
The other point is y=1, x=-2
The slope of one line is -2, (1-5)/2
The slope of the other is +2
The equation of the lines are y=-2x+5; y= 2x +5
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
I did this one yesterday.
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