   Chapter 2.P, Problem 11P

Chapter
Section
Textbook Problem

# How many lines are tangent to both of the circles x 2 + y 2 = 4 and x 2 + ( y − 3 ) 2 = 1 ? At what points do these tangent lines touch the circles?

To determine

To find:

how many lines are tangent to both thecircles  x2+y2=4  and x2+y-32=1 and check at what points these tangent lines touch the circles.

Explanation

Graph of both circles and tangent lines are shown in the following figure:

1) Given:

x2+y2=4 and x2+y-32=1

2) Calculations:

Let C1 be the x2+y2=4& C2 be the x2+y-32=1

From above figure, the line y = 2 is tangent to both the circles C1 and C2 at the point (0, 2).

Let us use implicit differentiation to find the slope of the tangent line.

Differentiate equation of first circle with respect to x:

x2+y2=4

2x+2ydydx=0

Divide both sides by 2,

x+ydydx=0

Subtract x from both sides,

ydydx=-x

Divide both sides by y,

dydx=-xy …………………. (1)

Thus the slope of tangent to C1 at point (x, y) is

dydx=-xy  =0                            …………………….(2)

Similarly, differentiate equation of second circle with respect to x.

x2+y-32=1

2x+2y-3dydx=0

Divide both sides by 2,

x+y-3dydx=0

Subtract x from both sides,

y-3dydx=-x

Divide both sides by y-3,

dydx=-xy-3……………………….. (3)

Thus the slope of tangent to C2 at point (x, y) is

dydx=-xy - 3  =0                            ………………. (4)

Thus, from (2) and (4), we get slope of tangent line m = 0 for both circles at point (0, 2).

Now substitute m = 0 and point (x1, y1)=(0, 2) in point slope form  y-y1=m (x-x1)

y-2=0 (x-0)

y=2

Therefore, the equation of the tangent line is y = 2.

From the figure, tangent line L1 passing through the points (a, b) & (c, d**#x0029; and if such a line exist then it’s reflection about y axis is another tangent line L2.

Also, the slope of L1 is same at point (a, b) and (c, d).

To find slope of circle C1 at point (a, b), substitute x = a and y = b in

dydx=-xy

dydx=-xy= -ab ………………..from (1)

Similarly, to find slope of circle C2 at point (c, d), substitute x = c and y = d in

dydx=-xy - 3

dydx=-xy-3  = -cd-3……………from (3)

-ab= -cd - 3

d-3=bca

d =  bca+3                           …………. (5)

Substitute x = c and y = d in x2+y-32=1

Since (c, d) is a point on C2 then it becomes,

c2+d-32=1

Substitute d-3=bca in above step,

c2+d-32=1

c2+bca2=1

a2c2+  b2c2 =a2

Factor out c2 from left side of the equation, it becomes

c2(a2 +  b2) =a2

Now substitute a2+  b2=4  in above step, since (a, b) is a point on x2+y2=4.

4c2=a2

By taking square root of both sides,

a=2c

Substitute  a=2c  in (5)

d =  bc2c+3

d =  b2+3 ……………….. (6)

Now find y intercept b1 using point (a, b) and slope

m= -ab.

Use slope intercept form: y=mx+b1

Substitute = -ab, x = a and y = b, then solve for b1

b1=a2b+b ………………………(7)

Similarly, find y intercept b2 using point (c, d) and slope m= -cd - 3.

Use slope intercept form: y=mx+b2

Substitute = -cd - 3, x = c and y = d, then solve for b2

d=-c2d - 3+b2

b2= d + c2d - 3 ……………………

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