Find the equation of:

1. The tangent to the circle x^2 - 2x + y^2 = 15 at(1,4)

2.the normal to the circle x^2 + 4x + y^2 - 6y = 0 at (1,1)

3. The tangent to the circle with C(1,2), r= root 5 at (3,3)

4. The normal to the cirlcle x^2 - 4x + y^2 = 9 at (5,2)

5. Show that the point P(-3,2) lies on the circle (x-1)^2 + (y+2)^2 = 32. Find the tangent to the circle at P.

6. The line 3x - y = 3 is tangent to the circle with centre (5,-1) and radius r. Find the value of r

Guest Oct 15, 2018

#1**+1 **

1. The tangent to the circle x^2 - 2x + y^2 = 15 at(1,4)

Implicitly differentiating the function, we can find the slope of a tangent line to the circle

2x - 2 + 2yy' = 0

y' = [2 - 2x ] / [2y]

y' = [1 - x] / y

At (1,4)....the slope is

[1 - 1 ] / 4 = 0

So...the equation of the tangent line at (1,4) is

y = 4

Here's the graph : https://www.desmos.com/calculator/uqvstutnla

CPhill Oct 15, 2018

#2**+1 **

2.the normal to the circle x^2 + 4x + y^2 - 6y = 0 at (1,1)

The slope of a tangent line at any point is found as

2x + 4 + 2yy' - 6y' = 0

y' [ 2y - 6] = - [2x + 4]

y ' = - [2x + 4 ] / [2y - 6]

At (1,1), the slope is

-[2(1) + 4 ] / [ 2(1) - 6 ] = - 6 / -4 = 3/2

Since we want a normal line at this point...the slope of our line is -2/3

So...the equation of the normal line is

y = (-2/3) ( x - 1) + 1

y =(-2/3)x + 2/3 + 1

Here's the graph : https://www.desmos.com/calculator/1jrzfuhcxk

CPhill Oct 15, 2018

#3**+1 **

3. The tangent to the circle with C(1,2), r= root 5 at (3,3)

The slope between the center and the point is

[ 3 - 2] / [ 3 - 1] = 1/2

The tangent line will have a negative reciprocal slope = -2

So.....the equation of the tangent line is

y = -2 ( x -3) + 3

y = -2x + 6 + 3

y = -2x + 9

Here's the graph : https://www.desmos.com/calculator/ve37r2kubn

CPhill Oct 15, 2018

#4**+1 **

4. The normal to the cirlcle x^2 - 4x + y^2 = 9 at (5,2)

Let's complete the square to find the center

x^2 -4x + 4 + y^2 = 4 + 9

(x - 2)^2 + (y - 0)^2 = 13

The center is (2, 0)

The slope between this point and ( 5,2) is

[ 2- 0 ] / [ 5 - 2 ] = 2/3 = slope of the normal line

The equation of the normal line at (5,2) is

y = (2/3) (x - 5) + 2

y = (2/3)x - 10/3 + 2

y= (2/3)x - 4/3

Here's the graph : https://www.desmos.com/calculator/u8n7sc01uh

CPhill Oct 15, 2018