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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
avatar+90968 
+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

 

 

cool cool cool

CPhill  Oct 15, 2018
 #2
avatar+90968 
+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

 

 

cool cool cool

CPhill  Oct 15, 2018
 #3
avatar+90968 
+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

 

 

cool cool cool

CPhill  Oct 15, 2018
 #4
avatar+90968 
+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

 

 

cool cool cool

CPhill  Oct 15, 2018

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