Systems of Equations Involving Circles and Lines

Transcription

1 Name: Systems of Equations Involving Circles and Lines Date: In this lesson, we will be solving two new types of Systems of Equations. Systems of Equations Involving a Circle and a Line Solving a system with a circle and a line is very similar to solving a linear system. While a few different methods can be used to solve a linear system, we will be focusing on the Substitution Method in this case. Our goal is to identify the ordered pairs which mark any intersection of a given circle and a line. Here are three types of situations. Figure 1 Figure Figure 3 Figure 1 shows the case where a line intersects a circle at two points. Because each intersection is a solution, this case has two solutions. Figure shows the case where the line is tangent to the circle. A line tangent to a circle intersects at only one point. Therefore, this case has one solution. Figure 3 shows the case where a line and circle do not intersect at all. Because no intersection eists, this case has no solution. Regardless of the case, all three situations can be approached in the same way algebraically using an adapted Substitution Method. The Substitution Method for a System Involving a Circle and a Line Step 1: Solve the linear equation for y. Solving for y is the same as getting the equation into slope-intercept form. Note: If the equation is of the form, = a, substitute the value of into the circle equation to find your point(s). Step : Substitute the value for y into the circle equation and solve for. Use either factoring or the quadratic formula to solve. The quadratic formula is: Given a + b + c = 0, b± b ac = for all values a 0. a Step 3: Substitute the value(s) for back into the linear equation and solve for y. Step : Check your solution(s) by substituting your point(s) in both given equations. 1

2 Eample 1 Solve the following system of a circle and a line by finding the intersection points algebraically. y = y = 100 Step 1: Solve the linear equation for y. This step is already complete since the linear equation was given in slope-intercept form. y = 3 30 Step : Substitute the value for y into the circle equation and solve for. We need to replace the y in our circle equation with 3 30 then move everything to the left-hand side of the equation. and y = (3 30) = (3 30)(3 30) = = = + = = 0 Substitute 3 30 for y. Epand (3 30). Combine like terms. Subtract 100 from both sides. Divide both sides by 10. We can now solve the remaining trinomial for all possible values of by either factoring or using the quadratic formula. Hint: When solving for, try factoring first. The Quadratic Formula can always be used if needed, but if the equation is factorable, you might be able to find the solution faster. In this eample, however, both methods will be shown. Factoring: = 0 ( 10)( 8) = 0 = 10 or = 8 Factor the left side of the equation. Zero-Product Property.

3 Quadratic Formula: a = 1, b = 18, and c = 80 ± ± = = (1) ( 18) ( 18) (1)(80) ± 18 ± = = = 9 ± 1 9± 1 gives us two values; = 10 and = 8. Step 3: Substitute the values for back into the linear equation and solve for y. When = 10 When = 8 y = 3 30 = 3(10) 30 = = 0 and y = 3 30 = 3(8) 30 = 30 = 6 We now have solutions, ( 10,0 ) and ( 8, 6). Step : Check your solutions by substituting your points in both given equations. For y = 3 30 For + y = 100 (0) = 3(10) 30 0 = = 0 + = (10) (0) = = 100 ( 6) = 3(8) 30 6= 30 6= 6 + = (8) ( 6) = = 100 Checking verified both points are correct solutions. Recall that Figures 1,, and 3 showed there are three possible scenarios for these types of problems. Because there were two solutions in Eample 1, the circle and the line intersect in two places as shown in Figure 1. 3

4 Systems of Equations Involving Two Circles Solving a system with two circles is essentially the same process as a system involving a circle and a line. This scenario is slightly more challenging, however, since it involves more squared variables. Let s look at the types of situations involving two circles. Figure Figure 5 Figure 6 Figure 7 Figure shows the case where the circles intersect at two points. Therefore, like previously, this case has two solutions. Figure 5 shows the case where one circle is tangent to the other and, therefore, only intersects at one point. This means that this case has one solution. Figure 6 shows a case like that in Figure 3. Here, neither circle touches each other. Because no intersection eists, this case has no solution. Figure 7 shows a unique case where the two equations graph the same circle. Since the circles intersect at every point, this case has infinite solutions. All four situations can also be approached algebraically in the same manner, this time using an adapted Elimination Method. The Elimination Method for a System Involving Two Circles Step 1: Multiply out all squared binomials in both equations. Then, move all nonvariable terms to the right side of the equation. Make sure the terms are in the same order for both equations. Step : Line up both equations (one on top of the other, with like terms under each other) and subtract. This will eliminate the and y terms. You will be left with a linear equation. Step 3: Solve the linear equation for y and substitute that value back into either circle equation. Solve the circle equation you chose for. You may get solutions. Note: The linear equation might be of the form = a or y = b. If that is the case, substitute that value into the correct variable of the equation you choose. Step : Substitute all values into the other circle equation. Solve that equation to find solutions for y. Note: This step might give more possible solutions than really eist. To test, make sure the points you found work in the linear equation from Step. Step 5: Check your solution(s) by substituting your point(s) in both given equations.

5 Eample Solve the following system of two circles by finding the intersection points algebraically. + y = ( ) ( ) + + y = Step 1: Multiply out all squared binomials in both equations. Then, move all non-variable terms to the right side of the equation. Make sure the terms are in the same order for both equations. The first equation is does not have any squared binomials and the non-variable term,, is already on the right side of the equation. We only need to focus on the second. ( ) ( ) + + y = = y y = y y + + = y y Epand ( + ) and ( y ). Combine like terms. Subtract 8 from both sides. Step : Line up both equations (one on top of the other, with like terms under each other) and subtract. Hint: Put in 0 place values to help when subtracting each term. ( y + 0y = ) ( + + y y = ) + y = 8 We are left with the linear equation + y = 8. This can be simplified since all terms are divisible by, leaving us with + y = Step 3: Solve the linear equation for y and substitute that value back into either circle equation. Solve the circle equation you chose for. + y = y = + Add to both sides. 5

6 Now let s substitute our value into one of the given circle equations. + y = + ( + ) = = + + = + = 0 + = 0 Substitute ( + ) for y. Epand ( + ). Combine like terms. Subtract from both sides. Divide both sides by. Now let s solve the equation for. + = 0 ( + ) = 0 = 0 or = Factor the left side of the equation. Zero-Product Property. Step : Substitute all values into the other circle equation. Solve that equation to find solutions for y. For = 0 For = ( + ) + ( y ) = ( + ) + ( y ) = (0 + ) + ( y ) = ( + ) + ( y ) = + y y + = y y + 8= y y + = 0 ( y )( y ) = 0 0+ y y + = y y + = y y = 0 yy ( ) = 0 y = y = 0 or y = This gives us three possible solutions: ( 0, ), (,0) and (, ). Let s check them with the linear equation from Step to avoid etra solutions. + y = + y = + y = (0) + () = ( ) + (0) = ( ) + () = = = 6 We can see the last point, (, ), is an etra solution, because unlike the first two, it does not make the linear equation true. Therefore, our solutions are: ( 0,) and (,0). 6

7 Step 5: Check your solution(s) by substituting your point(s) in both given equations. Although we checked ( 0,) and (,0) in the linear equation we found, we need to check our points in the given circle equations to verify accuracy. For + y = For ( ) ( ) + + y = + = (0) () 0+ = = (0 + ) + ( ) = + = () (0) + 0= = + = ( ) (0) + 0= = ( + ) + (0 ) = + = (0) ( ) 0+ = = Checking verified both points are correct solutions. Recall that Figures, 5, 6, and 7 showed there are four possible scenarios for these types of problems. Because there were two solutions in Eample, the two circles intersect in two places as shown in Figure. 7

8 Eercises Part A) Solve the following systems of a circle and a line by finding the intersection point(s), if any, algebraically. y = y = 5 + y = 15 ( ) + ( y 1) = 5 y = ( + ) + y = 3. Part B) Solve the following systems of two circles by finding the intersection point(s), if any, algebraically. + y = ( 3) + ( y + 3) = 9 + = + = ( 1) y 9 ( ) y 6. + = ( ) y 8 + y = 1 8