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Circle Equation: Identify Center and Radius (Examples)

Learn how to identify the center and radius of a circle from its equation through step-by-step examples.

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Circle Equation: Identify Center and Radius (Examples)

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  1. Identify the center and radius. From the equation, the graph is a circle centered at the origin with radius r= 36 = 6. EXAMPLE 1 Graph an equation of a circle Graphy2 = – x2 + 36. Identify the radius of the circle. SOLUTION STEP 1 Rewrite the equationy2 = – x2 + 36in standard form asx2 + y2 = 36. STEP 2

  2. EXAMPLE 1 Graph an equation of a circle STEP 3 Draw the circle. First plot several convenient points that are 6 units from the origin, such as (0, 6), (6, 0), (0, –6), and (–6, 0). Then draw the circle that passes through the points.

  3. The radius is = 29 r = (2 –0)2+ (–5 –0)2 = 4 + 25 29 EXAMPLE 2 Write an equation of a circle The point (2, –5) lies on a circle whose center is the origin. Write the standard form of the equation of the circle. SOLUTION Because the point (2, –5) lies on the circle, the circle’s radius rmust be the distance between the center (0, 0) and (2, –5). Use the distance formula.

  4. Use the standard form withr to write an equation of the circle. =29 = (29 )2 Substitute for r 29 EXAMPLE 2 Write an equation of a circle x2 + y2 = r2 Standard form x2 + y2 x2 + y2 = 29 Simplify

  5. A line tangent to a circle is perpendicular to the radius at the point of tangency. Because the radius to the point (1–3, 2)has slope = = – 2 – 0 2 – 3 – 0 3 EXAMPLE 3 Standardized Test Practice SOLUTION m

  6. y –2= (x –(–3)) y –2= x + y = x + 13 2 ANSWER 3 3 3 3 9 2 2 2 2 2 The correct answer is C. EXAMPLE 3 Standardized Test Practice the slope of the tangent line at (23, 2) is the negative reciprocal of or An equation of 2 – 3 the tangent line is as follows: Point-slope form Distributive property Solve for y.

  7. r = 9 = 3. for Examples 1, 2, and 3 GUIDED PRACTICE Graph the equation. Identify the radius of the circle. 1. x2 + y2 = 9 SOLUTION STEP 1 Explain is in the standard form x2 + y2 = 9 STEP 2 Identify the Center and radius form the equation, the graph is a circle centered at the origin with radius

  8. for Examples 1, 2, and 3 GUIDED PRACTICE STEP 2 Draw the circle. First plot several convenient points that are 3 units from the origin, such as (0, 3), (3, 0), (0, –3), and (–3, 0). Then draw the circle that passes through the points.

  9. 2. y2 = –x2 + 49 Identify the center and radius. From the equation, the graph is a circle centered at the origin with radius r= 49 = 7. for Examples 1, 2, and 3 GUIDED PRACTICE SOLUTION STEP 1 Rewrite the equationy2 = –x2 + 49in standard form asx2 + y2 = 49. STEP 2

  10. for Examples 1, 2, and 3 GUIDED PRACTICE STEP 3 Draw the circle. First plot several convenient points that are 7 units from the origin, such as (0, 7), (7, 0), (0, –7), and (–7, 0). Then draw the circle that passes through the points.

  11. r= 18 = 3 2. for Examples 1, 2, and 3 GUIDED PRACTICE 3. x2 – 18 = –y2 SOLUTION STEP 1 Rewrite the equation x2 – 18 = –y2in standard form asx2 + y2 = 18. STEP 2 Identify the center and radius. From the equation, the graph is a circle centered at the origin with radius

  12. for Examples 1, 2, and 3 GUIDED PRACTICE STEP 3 Draw the circle. First plot several convenient points that are 3 2 units from the origin, such as (0, 3 2 ), (3 2 , 0), (0, –3 2 ), and (–3 2 , 0). Then draw the circle that passes through the points.

  13. The radius is 26 for Examples 1, 2, and 3 GUIDED PRACTICE 4. Write the standard form of the equation of the circle that passes through (5, –1) and whose center is the origin. SOLUTION Because the point (5, –1) lies on the circle, the circle’s radius r must be the distance between the center (0, 0) and (5, –1). Use the distance formula. r= (5 – 0)2 + (–1 – 0)2 = 26

  14. Use the standard form withrto write an equation of the circle. = 26 Substitute for r x2 + y2 = ( 26)2 26 for Examples 1, 2, and 3 GUIDED PRACTICE x2 + y2 = r2 Standard form x2 + y2 = 26 Simplify

  15. 1 – 0 m = = 6 – 0 the slope of the tangent line at (6, 1) is the negative reciprocal of or –6 An equation of 1 6 1 6 for Examples 1, 2, and 3 GUIDED PRACTICE 5. Write an equation of the line tangent to the circle x2 + y2=37 at (6, 1). SOLUTION A line tangent to a circle is perpendicular to the radius at the point of tangency. Because the radius to the point (6, 1)has slope the tangent line is as follows: Point-slope form y –1= –6(x – 6)

  16. for Examples 1, 2, and 3 GUIDED PRACTICE y –1= –6x + 36 Distributive property Solve for y. y = –6x + 37

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