#### 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.

#### Answer

There are three lines y=-22x+6, y=22x+6 and y=2 tangent to both the circles x2+y2=4 and x2+y-32=1.

The line y=-22x+6 touches circle x2+y2=4 at point

423,23 and

x2+y-32=1 at point

(223,103)

The line y=22x+6 touches circle x2+y2=4 at point

-423,23 and x2+y-32=1 at point

(-223,103)

Also, the line y=2 touches both circles at point (0, 2).

#### 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 C_{1} be the x2+y2=4& C_{2} be the x2+y-32=1

From above figure, the line y = 2 is tangent to both the circles C_{1} and C_{2} 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 C_{1} 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 C_{2} 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 L_{1} 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 L_{2}.

Also, the slope of L_{1} is same at point (a, b) and (c, d).

To find slope of circle C_{1} at point (a, b), substitute x = a and y = b in

dydx=-xy

dydx=-xy= -ab ………………..from (1)

Similarly, to find slope of circle C_{2} 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 C_{2} 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 …………………….(8)

Now equate (7) and (8) since y intercepts are equal.

b1=b2

a2b+b= d + c2d - 3

Now substitute c=a2, d=3+ b2&d-3= b2 in above step,

a2b+b=3+ b2+ a24b2

Simplify right side of the equation, it becomes

a2b+b=3+ b2+ a24b2

a2b+b=3+ b2+ a22b

Multiply both sides by 2b, it becomes

2a2+2b2=6b+b2+a2

Subtract a2 and b2 from both sides,

a2+b2=6b

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

4= 6b

Divide both sides by 6,

b= 46= 23

Now find value of d, substitute b = 2/3 in d=3+ b2

d=3+ 232 = 103

To find value of a, substitute b = 2/3 in a2+b2=4

a2+232=4

a2=4- 232=329

a=423

To find value of c

c=a2

c=4232= 223

Therefore,

a, b=423,23&c, d=223,103

By using point

a, b=423,23

and point slope form

y-y1=mx-x1

y-23=-42323x-423

y-23=-22x-423

By using distributing property,

y-23=-22 x+163

Add 23 on both sides of the equation,

y=-22 x+6 Its reflection about y axis has equal to y=22 x+6

Therefore, there are three lines y=-22x+6, y=22x+6 and y=2 are tangents to both the circles x2+y2=4 and x2+y-32=1.

The line y=-22x+6 touches circle x2+y2=4 at point

423,23

and x2+y-32=1 at point

(223,103)

The line y=22x+6 touches circle x2+y2=4 at point

-423,23

and x2+y-32=1 at point

(-223,103)

Also, the line y=2 intersect both circles at point (0,2).

**Conclusion:**

There are three lines y=-22x+6, y=22x+6 and y=2 tangent to both the circles x2+y2=4 and x2+y-32=1.

The line y=-22x+6 touches circle x2+y2=4 at point

423,23 and

x2+y-32=1 at point

(223,103)

The line y=22x+6 touches circle x2+y2=4 at point

-423,23 and x2+y-32=1 at point

(-223,103)

Also, the line y=2 intersects both the circles at point (0, 2).