Find the equations of the tangents to the following circles at the points indicated 2.) x^2 +y^2 =25 , at (3,4) ; at (4,-3) ; at (0,5).

Answer 1

Since it says a geometry question so I solved it just with the basic algebra and without using derivatives.

All three points are on the circle since plugging the coordinates will satisfy the equation of the circle, for example

X^2 + Y^2 = 25, plug x =3 and y = 4 (lets call it point A) so 3^2 + 4^2 = 25 and then 16 + 9 = 25. The same is true for the other 2 points.

Now since the tangent is perpendicular to the radius at the point of tangency so the slope of OA is 4/3 and the slope of the tangent line at A is -3/4. Now just write equation of a line passing through A(3,4) with the slope of -3/4.

Y - Y1 = m(X - X1)

Y - 4 = -3/4( X - 3)

Y = -3 /4 X + 9/4 + 4 So, Y = -3/4 X + 25/4

With similar approach the equation of the tangent at (4 , -3) is

Y + 3 = 4/3( X -4) so Y = 4/3 X - 25/3

and the last point C (0,5) the slope of OC is infinity (5 / 0) so the tangent line from C is horizontal with the slope 0 so the equation of the tangent line is Y = 5.

Answer 2

(0,5) is the top of a circle with radius 5 and center at the origin. the tangent line is y=5

x^2 + y^2 =25

y^2 = 25-x^2

2yy' = -2x

y'=-x/y for (0,5) y'= -0/5 =0 slope of the tangent line = m

y=mx +b is the general form for a linear equation in slope intercept form

y=(0)x + b

now plug in the point (0,5)

5 = (0)(0) = b

b=5

the tangent line is y=5 a the point (0,5)

for (3,4) y' = -3/4 = slope

tangent line is

y=-3x/4 + b to solve for b plug in the point x=3, y=4

4 =-3(3)/4 + b

b =4 + 9/4 = 25/4

the tangent line is y=-3x/4 + 25/4

for (4,-3) y'= -4/-3 = 4/3

y=4x/3 + b

-3 =4(4)/3 + b

b = -3 -16/3 = -25/3

the tangent line is y=4x/3 -25/3