find the equation of a circle center at (-3,1)and tangent to a circle (x-1)^2+(y+2)^2=1.if the tangency Is:

Question

Both the two circles are tangent to each other .

Answer 1

The two centerpoints are (-3, 1) and (1, -2).  We want to find the equation of the line that passes through the two centerpoints.

 

slope = (-2 - 1) / (1 + 3)

         = -3 / 4

 

Using the point-slope form of a line,

 

y = mx - mx₁ + y₁

 

y = (-3/4)x - (-3/4)(-3) + 1

 

y = (-3/4)x - (9/4) + 1

 

y = (-3/4)x - (5/4)

 

 

Now this line will intersect the given circle.  Find the point of intersection. 

 

By rearranging the given circle equation, we make it easier to perform the algebra.

 

(y + 2)² = 1 - (x - 1)²

 

y + 2 = √(1 - (x² - 2x + 1))

 

y + 2 = √(2x - x²)

 

 

Substitute the line equation into this circle equation.

 

(-3/4)x - (5/4) + 2 = √(2x - x²)

 

(-3/4)x + (3/4) = √(2x - x²)

 

(-3/4)(x - 1) =  √(2x - x²)

 

 

Square both sides of the equation.

 

(9/16)(x² - 2x + 1) = 2x - x²

 

 

Multiply both sides of the equation by 16.

 

9x² - 18x + 9 = 32x - 16x²

 

 

Move all terms to the left side of equation.

 

25x² - 50x + 9 = 0

 

(5x - 1)(5x - 9) = 0

 

x = 1/5              or          x = 9/5

 

Since the point of intersection needs to between x=-3 and x=1,   we accept x=1/5

 

 

Evaluate the line equation at this x value.

 

y = (-3/4)(1/5) - (5/4)

 

y = (-3/20) - (5/4)

 

y = -28/20

 

y = -7/5

 

 

The point of intersection is (1/5, -7/5)

 

Now, find the distance between (-3, 1) and (1/5 , -7/5)

 

 

d = √[(-3 - (0.2))²+ (1 - (-1.4))²]

d = √[(-3.2)² + (2.4)²]

d = √(10.24 + 5.76)

d = √16

d = 4

 

This distance is the radius of the circle with centerponit (-3, 1)

 

The equation of the tangent circle is   (x + 3)² + (y - 1)² = 16