Contents. 2 Lines and Circles Cartesian Coordinates Distance and Midpoint Formulas Lines Circles...

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1 Contents Lines and Circles 3.1 Cartesian Coordinates Distance and Midpoint Formulas Lines Circles

2 CONTENTS

3 Chapter Lines and Circles.1 Cartesian Coordinates. Distance and Midpoint Formulas.3 Lines.4 Circles Definition.4.1. A circle is the set of points in the plane that are equidistant from a fixed point called the center of the circle. Each line segment with the center for one endpoint and a point on the circle for the other endpoint is called a radius of the circle. A segment with endpoints on the circle is called a chord of the circle. A chord that contains the center is caled a diameter. As with lines, we wish to know which equations have circles for graphs. answer is quite simple. The The Standard Form of the Equation of a Circle (x h) + (y k) = r (h, k) = center r = length of a radius As with lines, we need to show that every circle has an equation of the above form and that the graph of any equation of the above form has a graph that is a circle. To prove these facts, however, requires only a simple application of the distance formula. 3

4 4 CHAPTER. LINES AND CIRCLES x,y h,k Theorem.4.. The graph of the equation (x h) + (y k) = r is a circle with center (h, k) and radius of length r. Proof. The point (x, y) satisfies the equation (x h) + (y k) = r r = (x h) + (y k) d((x, y), (h, k)) = r (x, y) lies on the circle with center (h, k) and radius r. Example.4.3. Find an equation for the circle with center (, 3) passing through the point (3, 5). Sketch the circle. The center of the circle is given so (h, k) = (3, 3) so we can begin to write the equation: (x h) + (y k) = r (x ) + (y + 3) = r

5 .4. CIRCLES 5 y To complete the equation we need to find r. Using the distance formula: ,5 d = (x x 1 ) + (y y ) d = (3 ) + (5 + 3) = 65 r = 65 so r = 65. (x ) + (y + 3) = , x 10 In some cases an equation for a circle is given, but not in standard form. Such equations can be put in standard form by completing the square. Example.4.4. Find the center and radius of the circle with equation x + y 8 x + 6 y + 5 = 0. x 8 x + y + 6 y = 5 In order to complete the square we gather the x terms and the y terms together leaving spaces. x 8 x y + 6 y + 9 = Adding ( 8/) = 16 and (6/) = 9 to both sides. (x 4) + (y + 3) = center: (4, 3), r = Example.4.5. Find an equation for the circle with center (1, ) that is tangent to the line with equation x + y = 10. Sketch the circle. Again we are given the center of the circle so we can begin to write the equation: (x h) + (y k) = r

6 6 CHAPTER. LINES AND CIRCLES (x 1) + (y ) = r To find the radius in this case will require a bit more work. Our plan is this: 1. Write an equation for the line passing through the center perpendicular to the tangent. 8 y 6 4. Find the intersection point of these two lines. This intersection point is an endpoint of the radius drawn to the point of tangency. 1, x 3. Use the distance formula to find r. The tangent line is given by x + y = 10 which can be rewritten as y = 1 x + 5, hence m = 1, and m =. To find an equation of the line perpendicular to the tangent through the center we will use the point-slope form. y y 0 = m (x x 0 ) y = (x 1) y = x To find the intersection point we solve the system to the reader to check that the solution is (, 4). x + y = 10 y = x. We leave it

7 .4. CIRCLES 7 y 8 We now use the distance formula to find the length of a radius. 6 d = (x x 1 ) + (y y 1 ) 4 d = ( 1) + (4 ) = 5 Hence r = 5 and r = 5 1, x (x 1) + (y ) = 5 Theorem.4.6. The three perpendicular bisectors of the sides of a triangle are concurrent in a point which is equidistant from the three vertices of the triangle. The circle circumscribed about the triangle has this point as its center. Proof. C M 3 P M In A B C let M 1, M ans M 3 be the midpoints of sides A B, B C and A C respectively and let P be the point of intersection of the perpendicular bisectors through M 1 and M. A M 1 B In AP B, P M 1 is both an altitude and a median to side A B, hence A P B is isosceles with P A = P B. Similarly we conclude that P C = P B. Hence P A = P B = P C, so P is equidistant from the three vertices. To see that P lies on the

8 8 CHAPTER. LINES AND CIRCLES perpendicular bisector of the side A C we observe that since P A = P C, A P C is isosceles hence the median drawn to side A C is also an altitude. Now since P is equidistant from all three vertices, the circle with center P, and radius P C circumscribes A B C. Corollary.4.7. There is one and only one circle containing any three non-collinear points. Example.4.8. Find an equation for the circle circumscribed about the triangle with vertices A(0, 0), B(3, 0) and C(, 3). By.4.6 the center of this circle is the point common to the three perpendicular bisectors of the sides of the triangle, and the segments joining this point and the vertices are all radii of the circumscribing circle. Our plan for solution is:,3 1. Write equations for the perpendicular bisectors of all three sides.. Find the intersection point of two of the three lines. (The third necesarily intersects these two in the same point.) This is the center. 3. Use the distance formula to find the distance from the center to a vertex. This is the length of a radius. 1, 3 P 5, 3 0,0 3, 0 3,0 Using the midpoint formula, M = ( x1 + x, y ) 1 + y, we find that M 1 =

9 .4. CIRCLES 9 ( ) ( 3 5, 0, M =, 3 ) ( and M 3 = 1, 3 ) where M 1, M and M 3 are the midpoints of the sides A B, B C and A C respectively. Using the slope formula, m = y y 1, we find that m 1 = 0, m = 3, and m 3 = x x 1 3 where m 1, m and m 3 are the slopes of the sides A B, B C and A C respectively. Hence (m 1 ) is undefined, (m ) = 1 3, and (m 3) = 3 where (m 1), (m ) and (m 3 ) are the slopes of the perpendicular bisectors of the sides A B, B C and A C respectively. In the case of side A B, we observe that the equation for the perpendicular bisector is x = 3. For the other two sides we use the point-slope form, y y 0 = m (x x 0 ), to find the equations of the perpendicular bisectors. The equations are y = 1 3 x + 3 for side B C and y = 3 x for side A C. x = 3 Solving the system y = 1 3 x + we find (x, y) = ( 3, 6) 7. This is the center 3 of the circle. We use the distance formula d = (x x 1 ) + (y y 1 ), to find the length of the radius: r = d (( 3, ) ) , (0, 0) =. 6 We now use the standard form of the equation of the circle to find the equation of the circumscribing circle: (x h) + (y k) = r ( y 7 6 ( x 3 ) + ) = = 65 18

10 10 CHAPTER. LINES AND CIRCLES Problems

11 .4. CIRCLES 11 For problems 1 16 find an equation for and sketch each circle described. 1. center: (, 5), passing through ( 1, 0).. center : ( 3, 6), passing through (1, 7). 3. center : ( 3/, /3), passing through (3/4, 7/5). 4. center: (, ), tangent to the x axis. 5. center: (, 5), tangent to the y axis. 6. center: ( /3, 1/), tangent to the y axis. 7. center lies on the line y = x + 3 and tangent to both axes. (There are two such circles.) 8. center: (, 6), tangent to the line with equation x + 3 y = center: (5, 1), tangent to the line with equation x + 5 y = center: (5/3, 1/3), tangent to the line with equation x + 5 y = The cricle with the segment joining the points (3, 5) and (, 6) for a daimeter. 1. The circle with the segment joining the points (, 5) and (1, 7) for a daimeter. 13. The circle with the segment joining the points (1/3, 5/) and (1, 1/) for a daimeter. 14. The circle circumscribed about the triangle with vertices (1, 1), (3, 5) and (7, 0). 15. The circle circumscribed about the triangle with vertices (, 4), (, ) and (6, 8). 16. The circle circumscribed ( about the triangle with vertices, ) (, ) 3, 1 and (0, 1). 17. Find the center and radius of the circle with equation x + y 6x + 4y = Find the center and radius of the circle with equation 9x + 9y 9x 6y = Find the center and radius of the circle with equation 9x +9y +54x 1y = 58. For Problems 0 6 find the points of intersection between the circle and line whose equations are given. x + y = 5 0. x + y = 1 (x 3) + (y 1) = x y = 6 (x + 1) + y = 1. y = x 1 x + y = 5 3. x + y = 1 (x ) + (y + 1) = 3 4. x y = 4 x + y = 1 5. x + y = 1 (x 1) + y = 5 6. x + y = 1 7. Find an equation for the line tangent to the circle (x ) + (y 5) = 10 at the point (3, 8). 8. Find an equation for the line tangent to the circle x + y + 4 x 7 = 0 at the point (1, ). 9. Find an equation for the line tangent to the ( circle x + y ) + 4 x = 1 at the point., 30. Find an equation for the line tangent to the ( circle x + y ) + 4 x = 1 at the point.,

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