For what values of k will the line y=x+k be tangent to the circle x^2+y^2=25

Question

 This question is on pg 351 of the principles of mathematics 10 the section is 6.5 Interpreting Quadratic Equation Roots the answer is suppose to be 7.07 or -7.07 but I don't know the steps to get there

Answer 1

Note that the given line has a slope of 1 (i.e. an angle of π/4), so it can be tangent to the circle at a point in quadrant II and at a point in quadrant IV, depending on the value of k.  So there are 2 parallel lines of slope 1 that are tangent to the circle of radius 5 centered at the origin.  If you draw a graph, as Linda suggested, you will see that the quadrant II tangent point must be at

 

x = -r cos(π/4) = -5(√2)/2

y = 5 sin(π/4) = 5(√2)/2 

 

Similarly, the quadrant IV tangent point must be at

 

x = r cos(π/4) = 5(√2)/2

y = -5 sin(π/4) = -5(√2)/2 

 

Now set

 

x + k = y 

 

at these 2 points, and solve for k.

 

Quadrant II:

 

5(√2)/2 = -5(√2)/2 + k_II

 

k_II = 5(√2) ≅ 7.071

 

Quadrant IV:

 

-5(√2)/2 = 5(√2)/2 + k_IV

 

k_IV = -5(√2) ≅ -7.071

 

There is another way to find k using calculus, but I do not think that was what your teacher had in mind.

Answer 2

A point on the line y = x+k can be expressed as (x,x+k).

 

The slope of the tangent line is 1.

 

At the point of tangency, the tangent line is perpendicular to the radius of the circle that goes through the point of tangency

 

slope of line through (0,0) and (x,x+k) is (x+k)/x = -1

 

                                                             x+k = -x    So, k = -2x

 

Point of tangency = (x, x+k) = (x, -x)

 

Since (x,-x) lies on the circle, we have x² + (-x)² = 25     2x² = 25

 

                                                                                     x² = 25/2

 

                                                                                     x = ±5√2/2

 

k = -2x = ±5√2

 

Answer 3

Here is a calculus approach.

 

You want the line to be tangent to the circle.  Since the derivative is the slope of the tangent line, we can set the derivative of the circle to be 1.  Also, the line and circle must have the same point of intersection.

 

Implicit derive the circle.

 

2x + 2yy' = 0

 

2yy' = -2x

 

y'(2y) = -2x

 

y' = -x/y

 

Set y' equal to 1, since the slope of the line tangent to the circle is 1.

 

1 = -x/y

 

y = -x

 

 

Plug in this value into the circle equation to solve for x.  From here, you will find your points of tangency.

 

x² + (-x)² = 25

 

2x² = 25

 

x = ±√(25 / 2)

 

x = ±5/√2

 

x = -5/√2             and               x = 5/√2

 

Your points of tangency occur at these two x values.  Next, we plug them into the circle equation to get our two y values.

 

y = ±√(25 - x²)

y = ±√(25 - (25/2))

y = ±√((50 - 25) / 2)

y = ±√(25/2)

y = ±5/√2

 

So you points of tangency are (-5/√2, 5/√2) and (5/√2 , -5/√2).

 

Plug in the values of any of these points into the lines equation to solve for b.

 

5/√2 = -5/√2 + b

 

10/√2 = b 

 

5√2 = b

 

 

Using the other point,

 

-5/√2 = 5/√2 + b

 

-10/√2 = b

 

-5√2 = b

Answer 4

Always, first graph the problem.