Show that the point P(7,7) is on the perpendicular bisector of the line segment joining A (6,-1) to B (0, 3). Also, verify that P is the same distance from A and B.

For what values of k is the line y=l tangent to the circle (x-3)^2 + (y+2)^2 = 36?

Thanks!

Finished :D.

Ok, for the first part, find the midpoint of the line AB. THen, find the slope .

Now take the perpendicular slope to AB (the negative reciprocal).

Now you have the midpoint, and the slope of the perpendicular bisector. Write the equation in slope intercept form, and of course, solve for b. Show that the point P is on that line.

I don't understand the second part.

For the second part, let's find the equation of the line y = k that is a tangent to the circle (x - 3)^2 + (y + 2)^2 = 36.

To find the values of k, we need to determine the points of intersection between the circle and the line y = k.

First, substitute y = k into the equation of the circle:
(x - 3)^2 + (k + 2)^2 = 36

Simplify the equation by expanding and rearranging:
x^2 - 6x + 9 + k^2 + 4k + 4 = 36
x^2 - 6x + k^2 + 4k - 23 = 0

For the line y = k to be a tangent to the circle, this quadratic equation must have exactly one real root. In other words, the discriminant (b^2 - 4ac) of the quadratic equation must be equal to zero.

In this case, the quadratic equation is:
x^2 - 6x + k^2 + 4k - 23 = 0
a = 1, b = -6, c = k^2 + 4k - 23

The discriminant of this quadratic equation is:
D = b^2 - 4ac
= (-6)^2 - 4(1)(k^2 + 4k - 23)
= 36 - 4(k^2 + 4k - 23)
= 36 - 4k^2 - 16k + 92
= -4k^2 - 16k + 128

Set the discriminant D equal to zero:
-4k^2 - 16k + 128 = 0

Now, solve this quadratic equation by factoring or applying the quadratic formula. The values of k that satisfy this equation will correspond to the slopes of lines that are tangent to the circle.

I hope this helps! Let me know if you have any further questions.