Introduction We have worked with linear equations and equations of circles. We have solved systems of equations, including systems involving linear equations.

IntroductionWe have worked with linear equations and equations of circles. We have solved systems of equations, including systems involving linear equations and equations of parabolas. Now we will bring together skills and concepts from these topics and apply them in this lesson to solve systems of linear equations and equations of circles.

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6.3.1: Solving Systems of Linear Equations and Circles

Key Concepts• A system of equations is a set of equations with the

same unknowns. If a two-equation system has a linear equation and a circle equation, then the system can have no real solutions, one real solution, or two real solutions.

• A solution is an ordered pair; its graphical representation is a point at which the line and circle intersect, as shown on the following slide.

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Key Concepts, continued

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Key Concepts, continued• Systems of equations can be solved by using

graphical and/or algebraic methods. The graphical method is sometimes useful only for estimating solutions. No matter which method is used to solve a system, the solution(s) should be checked algebraically by substitution.

• A system with two real solutions will have two points of intersection between the line and the circle.

• A system with one real solution will have exactly one point of intersection.

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Key Concepts, continued• A system with no real solutions will have two complex

solutions that cannot be graphed on the Cartesian coordinate plane.

• Remember that the standard form of an equation of a circle is (x – h)2 + (y – k)2 = r2, where (h, k) is the center and r is the radius.

• A quadratic equation of the form ax2 + bx + c = 0 can have no real solutions, one real solution, or two real solutions.

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Key Concepts, continued• Using substitution, a system of a linear equation and a

circle equation can be reduced to a single quadratic equation whose solutions lead to the solutions of the system.

• The quadratic formula states that if a quadratic equation of the form ax2 + bx + c = 0 has one or two real solutions, then those solutions are given by

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Key Concepts, continued• To factor a polynomial means to write it as a product

of two or more polynomials. Examples:• x2 + 6x + 9 = (x + 3)(x + 3) • x2 + 5x – 6 = (x + 6)(x – 1) • 3x2 + 10x = x(3x + 10)

Note: x is a polynomial with one term, so it is a monomial.

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Key Concepts, continued• The Zero Product Property states that if a product

equals 0, then at least one of its factors is 0. Examples: • If ab = 0, then a = 0 or b = 0.• If (x + 6)(x – 1) = 0, then x + 6 = 0 or x – 1 = 0.

• Every quadratic equation of the form ax2 + bx + c = 0 that has one or two real solutions can be solved by the quadratic formula; some can be solved more easily by factoring and using the Zero Product Property.

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Key Concepts, continued• The distance formula states that the distance between

points (x1, y1) and (x2, y2) is equal to

• The midpoint formula states that the midpoint of the line segment connecting points (x1, y1) and (x2, y2) is

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Common Errors/Misconceptions• writing an incorrect sign for the last term of the trinomial

that results from squaring a binomial • using a instead of 2a as the denominator when applying

the quadratic formula • neglecting to key in parentheses when using a

calculator to square a quantity • neglecting to multiply both sides of an equation when

multiplying to eliminate fractions • forgetting to find the complex solutions to a system with

no real solutions

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Guided Practice

Example 2Solve the system below. Check the solution(s), then graph the system on a graphing calculator.

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Guided Practice: Example 2, continued

1. Solve the linear equation for one of its variables in terms of the other.

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–4x + 3y = 25 Equation of the line

3y = 4x + 25Add 4x to both sides to isolate the y-term on the left side.

Divide both sides by 3 to solve for y in terms of x.


2. Use the result from step 1 to substitute for y in the circle equation.

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Equation of the circle

Substitute for y.

Square the binomial.

Multiply both sides by 9 to eliminate fractions.


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Combine like terms.

Subtract 225 from both sides to get 0 on one side.

Divide both sides by 25 to simplify.

Factor.

Set the factors equal to 0.

Solve.


3. Substitute the x-value from step 2 into the linear equation and find the corresponding y-value.

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Linear equation solved for y

Substitute –4 for x.

Simplify, then solve.


The solution to the system of equations is (–4, 3).

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4. Check the solution by substituting it into both original equations.

The solution checks. The solution of the system is (–4, 3).

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Equation of the line: Equation of the circle:

–4x + 3y = 25 x2 + y2 = 25

–4(–4) + 3(3) = 25 (–4)2 + 32 = 25

16 + 9 = 25 16 + 9 = 25

25 = 25 25 = 25


5. Graph the system on a graphing calculator. First, solve the circle equation for y to obtain functions that can be graphed.

The linear equation solved for y is18



Subtract x2 from both sides to isolate y2 on one side.

If a2 = b, then because a can be negative.

Guided Practice: Example 2, continuedGraph the linear function and the two functions represented by the circle.

On a TI-83/84:

Step 1: Press [Y=].

Step 2: At Y1, type in [(][4][÷][3][)][X, T, θ, n][+][25][÷][3].

Step 3: At Y2, type in [ ][25][–][X, T, θ, n][x2][)].

Step 4: At Y3, type in [(–)][ ][25][–][X, T, θ, n][x2][)].

Step 5: Press [WINDOW] to change the viewing window. 19


Guided Practice: Example 2, continuedStep 6: At Xmin, enter [(–)][12].

Step 7: At Xmax, enter [12].

Step 8: At Xscl, enter [1].

Step 9: At Ymin, enter [(–)][8].

Step 10: At Ymax, enter [8].

Step 11: At Yscl, enter [1].

Step 12: Press [GRAPH].

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Guided Practice: Example 2, continuedOn a TI-Nspire:

Step 1: Press the [home] key.

Step 2: Navigate to the graphs icon and press [enter].

Step 3: At the blinking cursor at the bottom of thescreen, type in [(][4][÷][3][)][x][+][25][÷][3]

and press [enter].

Step 4: Move the cursor over to the double arrowat the bottom left of the screen and press

the center key of the navigation pad to open another equation prompt.

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Guided Practice: Example 2, continuedStep 5: At the blinking cursor, type in [ ][25][–][x]

[x2] and press [enter].

Step 6: Move the cursor over to the double arrow at the bottom left of the screen and press the center key of the navigation pad to open another equation prompt.

Step 7: At the blinking cursor type in [(–)][ ][25][–][x][x2] and press [enter].

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Guided Practice

Example 3Determine whether the line with equation x = –5 intersects the circle centered at (–3, 1) with radius 4. If it does, then find the coordinates of the point(s) of intersection.

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1. Write the system of equations represented by the line and the circle.

The equation of the circle with center (h, k) and radius r is (x – h)2 + (y – k)2 = r2, so the equation of the circle is [x – (–3)]2 + (y – 1)2 = 42, or (x + 3)2 + (y – 1)2 = 16. The system is:

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Equation of the line



2. Substitute the x-value from the linear equation into the equation of the circle and then solve the resulting equation.

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(x + 3)2 + (y – 1)2 = 16 Equation of the circle

(–5 + 3)2 + (y – 1)2 = 16 Substitute –5 for x.

4 + (y – 1)2 = 16 Simplify.

4 + y2 – 2y + 1 = 16 Square the binomial.

y2 – 2y + 5 = 16 Combine like terms.

y2 – 2y – 11 = 0 Subtract 16 from both sides to get 0 on the right side.


Apply the quadratic formula. The solutions of

ax2 + bx + c = 0 are so the

solutions of an equation in the form ay2 + by + c = 0

are

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The solutions to the quadratic equation are

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Apply the quadratic formula.

Simplify.


3. The linear equation in the system is x = –5, so any solution must have –5 as its x-coordinate. Use this fact along with the results from step 2 to write the solutions of the system. The solutions of the system are

These are the coordinates of the points at which the line intersects the circle, as shown on the following graph.

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Introduction We have worked with linear equations and equations of circles. We have solved systems of equations, including systems involving linear equations.

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