A line is tangent to the circle x^2+y^2=41, at the point (-4,5)
find the slope of the radius from the center to the point of tangency
write the equation of the line tangent to the given circle at the given point

If the equation of the circle is x^2+ y^2 = 41, we must first understand the parts of the equation.
A general circle’s equation is (x-h)^2+(y-k)^2= r^2
(h. k) is the radius of the circle
r is the radius of the circle
Another useful fact to know is that tangent lines touch the circle at one point (4,5)

Since in our original equation there are no h or k values, we can assume that the center of the circle is (0,0).

The formula for slope is Y1-Y2
X1-X2
We can break this down with our two points (center and tangent)
(0,0)    and   (-4,5)
(X1, Y1) and (X2, Y2)

therefore, we will put the equation as such
0-(-5)= 5  =  5
0-(-4)= 4      4

this is our slope from the center to the point of tangency.

We know that tangent lines are perpendicular to the radius, which we’ve already found the slope of. Perpendicular lines are opposite reciprocals of the line they are perpendicular to.

Therefore, we take our slope from center to the tangent, and make it opposite and then take the reciprocal of that slope, which will give us the slope of the tangent line itself. (note: reciprocal means flip the numerator and denominator)
5 =  -5  =  -4
4     4        5

Now, we have a point on the line, and the line’s slope. We can use slope-intercept equation to find the equation of the line.

Slope-int      y=mx+b
(x, y) is a point,
m is the slope
b is the y intercept  ( the point where x=0, or where its on the y axis)

now we plug things in
(-4,5) is our point,
-4  is our slope
5

-5=-4(-4)+b       After we plug things in, solve for b
5

-5= 3.2+b

-1.8= b  or  b=  1 4
5

Now we just need to rewrite our equation with all our components.
(-4.5) = point
-4  = slope
5
1 4 = y-intercept
5

y=-4 x+ 1 4     This is the equation of the tangent line
5         5

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