Math  /  Geometry

QuestionThe diagram shows two circles that touch at D(6,3)D(6,3).
Circle C1C_{1} has equation (x2)2+(y5)2=20(x-2)^{2}+(y-5)^{2}=20. Circle C2C_{2} is twice the size of C1C_{1}. Find, (a) The equation of Circle C2\mathrm{C}_{2}. 5 (b) The equation of the common tangent at D . 3

Studdy Solution

STEP 1

What is this asking? We've got two circles that kiss at a point, and we know one circle's equation and that the other is *twice* its size.
We need to find the equation of the bigger circle and the equation of their shared tangent line at the kissing point. Watch out! "Twice the size" could mean twice the radius or twice the area.
We need to figure out which one it is!
Also, remember that a tangent line only touches a circle at one point and is perpendicular to the radius at that point.

STEP 2

1. Find the center and radius of C1C_1.
2. Find the center and radius of C2C_2.
3. Find the equation of C2C_2.
4. Find the equation of the tangent line.

STEP 3

The equation of C1C_1 is given by (x2)2+(y5)2=20(x-2)^2 + (y-5)^2 = 20.
Remember, the general equation of a circle is (xh)2+(yk)2=r2(x-h)^2 + (y-k)^2 = r^2, where (h,k)(h,k) is the center and rr is the radius.

STEP 4

Matching this with the given equation, the center of C1C_1 is (2,5)(2,5).
Awesome!

STEP 5

We also see that r2=20r^2 = 20, so the radius of C1C_1 is r=20=45=25r = \sqrt{20} = \sqrt{4 \cdot 5} = 2\sqrt{5}.
Perfect!

STEP 6

C2C_2 is *twice* the size of C1C_1.
Since the problem doesn't specify, let's assume "twice the size" refers to twice the *area*.

STEP 7

The area of C1C_1 is A1=π(25)2=20πA_1 = \pi (2\sqrt{5})^2 = 20\pi.
The area of C2C_2 is twice this, so A2=220π=40πA_2 = 2 \cdot 20\pi = 40\pi.

STEP 8

Let RR be the radius of C2C_2.
Then, A2=πR2=40πA_2 = \pi R^2 = 40\pi.
Dividing both sides by π\pi, we get R2=40R^2 = 40, so R=40=410=210R = \sqrt{40} = \sqrt{4 \cdot 10} = 2\sqrt{10}.

STEP 9

Since the circles touch at D(6,3)D(6,3), the centers of both circles and DD lie on the same line.
Think of it like two pearls on a string touching each other!

STEP 10

Let the center of C2C_2 be (h2,k2)(h_2, k_2).
The distance from the center of C1C_1 to DD is 252\sqrt{5}.
The distance from the center of C2C_2 to DD is 2102\sqrt{10}.
Since 210=2252\sqrt{10} = \sqrt{2} \cdot 2\sqrt{5}, the center of C2C_2 is 2\sqrt{2} times farther away from DD than the center of C1C_1.

STEP 11

The vector from the center of C1C_1 to DD is (62,35)=(4,2)(6-2, 3-5) = (4,-2).
Multiplying this vector by 2\sqrt{2} and adding it to DD gives us the center of C2C_2: (6+42,322)(6 + 4\sqrt{2}, 3 - 2\sqrt{2}).

STEP 12

We now have the center (6+42,322)(6 + 4\sqrt{2}, 3 - 2\sqrt{2}) and radius 2102\sqrt{10} of C2C_2.

STEP 13

Plugging these values into the general equation of a circle, we get the equation of C2C_2: (x(6+42))2+(y(322))2=(210)2(x - (6 + 4\sqrt{2}))^2 + (y - (3 - 2\sqrt{2}))^2 = (2\sqrt{10})^2, which simplifies to (x642)2+(y3+22)2=40(x - 6 - 4\sqrt{2})^2 + (y - 3 + 2\sqrt{2})^2 = 40.

STEP 14

The tangent line at DD is perpendicular to the radius connecting the center of C1C_1 to DD.

STEP 15

The slope of the radius of C1C_1 to DD is 3562=24=12\frac{3-5}{6-2} = \frac{-2}{4} = -\frac{1}{2}.

STEP 16

The slope of the tangent line is the negative reciprocal of the radius slope, which is 11/2=2-\frac{1}{-1/2} = 2.

STEP 17

Using the point-slope form with point D(6,3)D(6,3) and slope 22, the equation of the tangent line is y3=2(x6)y - 3 = 2(x - 6).

STEP 18

Simplifying, we get y3=2x12y - 3 = 2x - 12, or y=2x9y = 2x - 9.

STEP 19

(a) The equation of circle C2C_2 is (x642)2+(y3+22)2=40(x - 6 - 4\sqrt{2})^2 + (y - 3 + 2\sqrt{2})^2 = 40. (b) The equation of the common tangent at DD is y=2x9y = 2x - 9.

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