Find the equation of a line tangent to the circle x^2+y^2=8 at the point (–2, 2).

Question

PLEASE FIND THE EQUATION AND THANK YOU IF YOU DO!

Answer 1

Start by plotting this on graph paper. You have a circle centered at (0,0) with a radius of √8.

From the plot, you should be able to see that point (-2,2) is in the 2nd quadrant, and is at a 45 degree angle from each axis. It should be apparent from the graph that the slope of the tangent line at this point is +1.

For the equation, we can use the "slope-intercept" form: y = mx + b, where m is the slope and b is the y-intercept.

Plug in the value of m from above, and the coordinates of (-2,2):

2 = 1*(-2) + b

b = 4, so the equation is y = x + 4. Verify this is correct by plotting it on the same graph with the circle.

You can also use an online graphing tool such as DESMOS:

https://www.desmos.com/calculator

Answer 2

It's a circle with center at the origin, with radius square root of 8 or 2(2)^1/2

At the point (-2,2), x squared plus y squared = 4+4=8. It's definitely a point on the circle, but in the 2nd quadrant, top left part of the circle. If you sketch it, it seems to be a point that's at a 45 degree angle, the tangent line with slope of 1. Use the point slope formula y-k = m(x-h) The point (h,k) = (-2,2) and m=1, so the line is y-2=(1)(x+2) or y=x+4 The tangent line at that point is just the y=x diagonal line through the origin shifted up by 4 units.

Or if you're into calculus, calculate the slope by taking the first derivative of the circle equation, with implicit differentiation. First derivative gets you to 2x+2yy'=0 or y'=-x/y x=-2 y=2 so y'=-(-2)/2=1 tangent line has slope 1 or a 45 degree line going through (-2,2)