Notice that the center points of the drawn circles have y-coordinates of 0.
Equation: (x-15)^2+y^2=100 Explanation: See solution.
Practice makes perfect
We are given that four tangent circles are centered on the x- axis. If we call the radius of ∘ O r, then the radii of the consecutive circles will be 2 r,3 r and 4 r.
Since the circles are centered on the horizontal axis the center points of these circles have y-coordinates of 0. Notice that we can evaluate the x-coordinates of the center points by adding the radii of consecutive circles.
A( r+2 r, 0) & → A(3r,0)
B(3r+2 r+3 r, 0) & → B(8r,0)
C(8r+3 r+4 r, 0) & → C(15r,0)Now, using the coordinates of the center of ∘ C and its radius, we can write an equation using the standard equation of a circle.
(x-15r)^2+(y- 0)^2=(4 r)^2 ⇓
(x-15r)^2+y^2=16r^2
Since we know that the point (63,16) lies on ∘ C, we can substitute coordinates of this point into the equation and solve for r.
Next we will use the Quadratic Formula to solve for r.
ax^2+ bx+ c=0 ⇕ x=- b± sqrt(b^2-4 a c)/2 a
We first need to identify the values of a, b, and c.
209r^2-1890r+4225=0 ⇕ 209r^2+( -1890)r+ 4225=0
We see that a= 209, b= -1890, and c= 4225. Let's substitute these values into the Quadratic Formula.
The solutions for this equation are r= 1890± 200418. Let's separate them into the positive and negative cases.
r=1890± 200/418
r_1=1890+200/418
r_2=1890-200/418
r_1=2090/418
r_2=1690/418
r_1=5
r_2=4.043...
Since we are given that all circles have integer radiuses, the only possible solution is r= 5. Knowing this value. we can find the equation of ∘ A since we found before the center of this circle is A(3r,0) and the radius is 2r.
