Intersection of lines and 4. circles Touching Circles

How can we tell if two circles are r C2 2 touching? r1 C 1 uuuur When circles are touching: uuuur CC12 CC12 = r1 + r2

C2 r2

r1 When circles DO NOT touch: C1

is bigger than r1 + r2 Touching Circles

How can we tell if two circles are r C2 2 touching? r1 C 1 uuuur When circles overlap: uuuur CC12 CC12 is less than r1 + r2

The distance formula can be used to find Example By considering centres and radii prove that the following two circles touch each other. Circle 1 x2 + y2 + 4x - 2y - 5 = 0 Circle 2 x2 + y2 - 20x + 6y + 19 = 0

Solution: 22 C1(− 2,1) r1 =(2) + ( − 1) − ( − 5)

r1 = 10

22 C2 (10,− 3) r2 =( − 10) + (3) − 19

r2 = 90

r2 =9  10 = 3 10 Example By considering centres and radii prove that the following two circles touch each other. Circle 1 x2 + y2 + 4x - 2y - 5 = 0 Circle 2 x2 + y2 - 20x + 6y + 19 = 0

Solution: uuuur 22 C12 C= d =(10 − ( − 2)) + ( − 3 − 1) C1(− 2,1) r1 = 10 uuuur 22 C12 C= d =(12) + ( − 4) C2 (10,− 3) uuuur C C== d 160 r =9  10 = 3 10 12 2 uuuur C12 C= d =16  10 = 4 10 uuuur Since C 1 C 2 =+ r 1 r 2 the circles are touching Intersection of lines and circles

Copy the following:

There are 3 possible scenarios

2 points of contact 1 point of contact 0 points of contact ie line is a tangent

To determine where the line and circle meet we use simultaneous equations though the method is slightly different from that for intersecting lines and a line with a parabola. Example 1

Find the points of intersection of the line y = 4 and the circle x22+ y −4 x − 6 y + 8 = 0

Solution:

Sub y = 4 into equation for circle: x22+ y −4 x − 6 y + 8 = 0

xx22+(4) − 4 − 6(4) + 8 = 0 xx2 +16 − 4 − 24 + 8 = 0

2 Look to factorise: xx−=40 xx(−= 4) 0

x = 0 or x = 4 Points of contact are: (0, 4) and (4, 4) Example 2 Find where the line y = 2x + 6 meets the circle x2 + y2 + 10x – 2y + 1 = 0 .

Solution:

22 Sub y = 2x + 6 into equation for circle: x+ y +10 x − 2 y + 1 = 0

x22+(2 x + 6) + 10 x − 2(2 x + 6) + 1 = 0 x22+4 x + 24 x + 36 + 10 x − 4 x − 12 + 1 = 0

2 Look to factorise: Common factor of 5 5xx+ 30 + 25 = 0 xx2 +6 + 5 = 0 (xx+ 5)( + 1) = 0 x = -5 or x = -1 Example 2 Find where the line y = 2x + 6 meets the circle x2 + y2 + 10x – 2y + 1 = 0 .

Solution:

Find y coord by subbing these values into x = -5 or x = -1 equation for straight line

y =2( − 5) + 6 y =2( − 1) + 6 y =−4 y = 4

Points of contact are: (-5, -4) and (-1, 4) Example 3 TANGENCY NAB Prove that the line 2x + y = 19 is a tangent to the circle x2 + y2 - 6x + 4y - 32 = 0 , and also find the point of contact. Solution: NB. Tangency means that there is only one point of contact between line and circle so we are aiming at a quadratic with one (repeated) root and so b2 – 4ac = 0

Must have y = or x = yx=−19 2

Sub y = 19 - 2x into x22+(19 − 2 x ) − 6 x + 4(19 − 2 x ) − 32 = 0 equation for circle: x22+(361 − 76 x + 4 x ) − 6 x + 76 − 8 x − 32 = 0 5xx2 − 90 + 405 = 0 Example 3 TANGENCY NAB Prove that the line 2x + y = 19 is a tangent to the circle x2 + y2 - 6x + 4y - 32 = 0 , and also find the point of contact. Solution:

Look to factorise: Common factor of 5 5xx2 − 90 + 405 = 0

xx2 −18 + 81 = 0 Example 3 TANGENCY NAB Prove that the line 2x + y = 19 is a tangent to the circle x2 + y2 - 6x + 4y - 32 = 0 , and also find the point of contact.

Solution: xx2 −18 + 81 = 0 Method 1 (use if need to find POC) (xx− 9)( − 9) = 0

Find y coord by subbing this value into x = 9 (repeated) equation for straight line y = 19 - 2(9)

y = 1 There is only one point of contact at (9, 1) and so the line is a tangent to the circle. Example 3 TANGENCY NAB Prove that the line 2x + y = 19 is a tangent to the circle x2 + y2 - 6x + 4y - 32 = 0 , and also find the point of contact.

Solution: xx2 −18 + 81 = 0 Method 2 (use if only need to prove tangency) b2 − 4 ac =( − 18)2 − 4(1)(81) =−324 324 As discriminant is zero there is only = 0 one real root and so one point of contact and so the line is a tangent to the circle. Example 3 TANGENCY NAB Prove that the line 2x + y = 19 is a tangent to the circle x2 + y2 - 6x + 4y - 32 = 0 , and also find the point of contact.

Solution: xx2 −18 + 81 = 0 We could then find POC as follows: −b  b2 − 4 ac x = 2a Discriminant is Zero (18) 0 x ==9 repeated 2 Find y coord by subbing this value y = 19 - 2(9) Into equation for straight line y = 1 There is only one point of contact at (9, 1) and so the line is a tangent to the circle. Heinemann, p.218, EX 12K, Q1(a) to (e), Q 2, 3, & 6 (b) and (c)