﻿ Analytical geometry examples
Analytical geometry solved examples                 18 Circle given by two points and tangent to the y axis ▲
Given two points on a circle (6 , 2) and (3 , -1), the circle is tangent to the y axis. Find the radius and the center coordinate of the circle.
 Draw the circle and mark the point of the circle center that is at a = R and b (because the circle is tangent to the y axis, b is unknown) We can write the equation of the circle related to the points and the center of the circle:
 ( x1 − R )2 + ( y1 − b )2 = R2 (1) ( x2 − R )2 + ( y2 − b )2 = R2 (2)
 From first eq. (1) x12 − 2x1R + R2 + y12 − 2y1b + b2 = R2 x12 − 2x1R + y12 − 2y1b + b2 = 0 From first eq. (2) x22 − 2x2R + R2 + y22 − 2y2b + b2 = R2 x22 − 2x2R + y22 − 2y2b + b2 = 0
Substitute R from eq. (3) into the last equation and arrange terms:
b2(2x1 − 2x2) + b(4x2y1 − 4x1y2) + 2x1x22 − 2x12x2 + 2x1y22 − 2x2y12 = 0
b2(x1 − x2) + b(2x2y1 − 2x1y2) + x1x22 − x12x2 + x1y22 − x2y12 = 0
We get a quadratic equation with the unknown b (all the points are given) the solution is: Now insert the value of b into eq. (3) to get the radius R of the circle.
Insert the given values of the points (6 , 2) and (3 , − 1) to get b: The radius from eq. (3) for b = 2 is: And the second radius for b = − 10 is: We can see that there are two circles that fulfills the requirements of the given data.
The two circles centers are at the points (3 , 2) and (15 , − 10)
 19 Circle given by two points and the radius ▲
Given two points on a circle (2 , -1) and (-2 , 7) and the radius which is equal to 5. Find the center coordinate of the circle.
We can write the equation of the circle with two points which are located on the circle as:
 (x1 - a)2 + (y1 - b)2 = R2 (x2 - a)2 + (y2 - b)2 = R2
We get two equations with two unknowns  a  and  b  but to find their values from this equations is not an easy job, you can try it.
So we will try another method to solve the problem. Connect the two points so it make a lineThe mid point coordinate of this two points is: xm = (x1 + x2) / 2 ym = (y1 + y2) / 2
 The slope of the line connecting the two points is: Line AB is perpendicular to this line and the slope is: The equation of line AB based on point (xm , ym) is: y − ym = mAB (x − xm)
Line AB is passing through the center of the circle (point a,b) so now we search a point on this line which is at a distance of R fron the points (x1 , y1), so we have the equations
 Distance from point 1 to (a , b) is R: (1) Point (a , b) is on the line AB so: (2)
Substituting the value of  b  from eq. (2) into eq. (1) we get:
(x1 − a)2 + [y1 + mAB(xm − a) − ym]2 = R2
 In order to simplify the equation we will mark z as: z = y1 + mABxm − ym
x12 − 2x1a + a2 + (z − mABa)2 = R2
 And we get the quadratic equation: a2(1 + mAB2) − a(2x1 + 2zmAB) + x12 + z2 − R2 = 0 The value of b is found by substituting the values of a into eq. (2).
 According to given data:  z = − 1 + 0.5 * 0 − 3 = − 4  And   b   is according to eq. (2): b1 = 3 + 0.5(2 − 0) = 4 b2 = 3 + 0.5(− 2 − 0) = 2
So the two possible circles centers are at:       (2 , 4)   and   (− 2 , 2)
Point 1: ( , ) Point 2: ( , ) Radius:
 Circles centers:
 Two circles tangency - example 20 ▲ Determine the tangency point of two circles of the form
 (x − x1)2 + (y − y1)2 = r12 (x − x2)2 + (y − y2)2 = r22
Then solve numerically with the values:     y1 = y2 = 2     x1 = − 10
and x2 = 4     r1 = 5     r2 = 9

The condition for tangency are:
 Outer tangency: d = r1 + r2 Inner tangency: d = | r1 − r2 | The slope of the line that connectes circles centers  m  is: (1)
Because the tangency point is on this line we can use simple trigonometry to find the point location:
From the similarity of two trangles we have:
 For the x axis we have: ⟶ For y axis we have: ⟶ Equations of the circles are: (x + 10)2 + (y − 2)2 = 52 (x − 4)2 + (y − 2)2 = 92
 The tangency point is:   Determine the tangency point of two circles of the form
 (x − 1)2 + (y + 4)2 = 92 (x − 1)2 + (y − 1)2 = 42

The distance between circles centers is: Because     |r1 − r2| = |9 − 4| = 5 = d
then the circles are inner tangent.
Because the tangency point is on this line we can use simple trigonometry to find the point location:
From the similarity of two trangles we have:
 For the x axis we have: ⟶ For y axis we have: ⟶ The tangency point is:  