• ## Discover

1. Home >
2. All Categories >
3. Science & Mathematics >
4. Mathematics >
5. Resolved Question
Member since:
April 07, 2012
Total points:
92 (Level 1)

## Resolved Question

Show me another »

# Urgent, Pls help , a question of tangent?

Let C1 and C2 be circles defined by x^2 − 20x + y^2 + 64 = 0 and (x+15)2 + y2 = 81
respectively. What is the length of the shortest line segment PQ that is tangent to C1 at P
and to C2 at Q?

please explain the steps, and if you can, please show me the graph, thanks a lot ~
Member since:
November 01, 2009
Total points:
107,473 (Level 7)

On "Axes", change the range of X edge from - 25 to 20, then "OK".

On "Function I Insert relation", type x^2 − 20x + y^2 + 64 = 0, then "OK".

On "Function I Insert relation", type (x + 15)^2 + y^2 = 81, choose another color, then "OK".

### Source(s):

Graph 4.4
By the way, how to cal the ans then, i click cal the length of the path... then how to select the point ??? thanks a lot :) please ans me at http://answers.yahoo.com/question/index;_ylt=AsEuQTdT.LmUg5WdkA_y_Ujsy6IX;_ylv=3?qid=20120406235943AAUPHoy
Great thanks

There are currently no comments for this question.

This question about "Urgent, Pls help , a… " was originally asked on Yahoo Answers United States

• by M
Member since:
September 09, 2011
Total points:
12,490 (Level 6)
Show the graph here? Not sure how to do this. But we may try the mathematics approach.

C1: x^2 − 20x + y^2 + 64 = 0

=> (x - 10)^2 + y^2 = -64 + 10^2

=> (x - 10)^2 + y^2 = 6^2

Original of C1 is O1:(10, 0)

C2: (x+15)^2 + y^2 = 81

=> (x+15)^2 + y^2 = 9^2

Original of C1 is O2:(-15, 0)

line segment PQ that is tangent to C1 at P, and to C2 at Q?

O1 to P form a 90 degree (since P is tangent to C1); so is O2 to Q.

The distance between O1O2 is 10 - (-15) = 25

Now, next step you will need to use your imaginary to picture this,

draw a line from P to C1, and it will parallel to the x-axis.

Let say this line reach at line O2Q at point S. You may see the O1PSO2 forms a rectangle.

next, O2Q is the radius of C2 which is 10, and line O2S is 9. (Why? Because O1P and O2S are the same width of this rectangle you just formed.)

Therefore, QS = 1

For right triangle PQS, the hypotenuse will be PS and the two legs are PQ and QS.

PS^2 = PQ^2 + QS^2

PS is the length os O1O2 which is 25

25^2 = PQ^2 + 1^2

625 = PQ^2 + 1

PQ^2 = 624

PQ = 4√39