The line with equation y = x + k is a tangent to the circle x^2 + y^2 = 25. Find the values of k. Help please!

Question

scythian how did you get from:

x + k = √(25-x²) - I've got here..

to

x = (1/2)(- k ± √(50-k²))

Answer 1

Solve the simultaneous set of equations. That is, in this case, solve for x:

x + k = √(25-x²)

x = (1/2)(- k ± √(50-k²))

Since for a tangent point, there can only be one value for x, this means that √(50-k²) has to be 0, or k = ± 5√2. The reason why there are TWO values for k is because two possible tangent lines are possible, on opposite sides of the circle.

Answer 2

Scythina got there by simply squaring both sides and solving the resulting quadratic equation.

But it's way more simple than that. The line has slope 1 (45 degrees) and is tangent to the circle of radius 5 centered at the origin. The line and the circle will touch at the point (-5/sqrt2, 5/sqrt2), or at the opposite point on the circle (25/sqrt2, -25/sqrt2). Plugging these points in the line equation y=x+k, we get the two solutions:
5/sqrt2 = -5/sqrt2 + k → k = 5sqrt2
-5/sqrt2 = 5/sqrt2 + k → k = -5sqrt2

Answer 3

The slope of the tangent line is 1. The circle is centered at the origin with a radius of 5.

There are two possible answers. The line intersects the circle in quadrant II or IV at point (-3.535, 3.535) or (3.535, -3.535)

(y-3.535) = 1 (x+3.535) or (y+3.535) = 1 (x-3.535)

the Y intercept, K, will be either 7.071 or -7.071

Best answer please.

Answer 4

find the point of intersection of the line and the circle
then you will get a quadratic equation
then used b^2-4ac=0
then solve it u get your k value

try it because this question is so familiar in my school.