Sample Problems. Practice Problems

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1 Lecture Notes Circles - Part page Sample Problems. Find an equation for the circle centered at (; ) with radius r = units.. Graph the equation + + = ( ).. Consider the circle ( ) + ( + ) =. Find all points on the circle with coordinate being.. Find the coordinates of all points where the circle + ( ) = and the line = + 8. a) Consider the circle + =. Find an equation for the tangent line drawn to the circle at the point (; ). b) Consider the circle + + =. Find an equation for the tangent line drawn to the circle at the point (; ).. a) Find the points where the circles ( + ) + ( + ) = and ( + ) + ( ) = b) Find the points where the circles ( + ) + ( + ) = and ( ) + ( ) = Practice Problems. Find an equation for each of the following circles. C will denote the center and r the radius. a) C (; ) r = 7 b) C (; ) r = c) C ( 8; ) r = p. Graph the equation + + = ( + ). Consider the circle ( + 7) + ( ) =. Find all points on the circle a) with coordinate. c) with coordinate. b) with coordinate. d) with coordinate.. Find the coordinates of all point(s) where the circle and the line a) ( + ) + ( 7) = 8 and + = c) + ( + ) = and = b) ( ) + ( + ) = and = + 9 d) ( ) + ( + ) = and = +. Given the equation of a circle and a point P on it, nd an equation for the tangent line drawn to the circle at the point P. a) + = and P ( 8; ) b) ( ) + ( + ) = and P (; ). Find the coordinates of all points where the given circles a) ( ) + ( ) = and ( ) + ( ) = b) ( ) + = and + ( ) = c) + ( 8) = and ( 7) + ( ) = c copright Hidegkuti, Powell, 8 Last revised: March 9,

2 Lecture Notes Circles - Part page Sample Problems - Answers.) ( ) + ( + ) =.) ( + ) + ( ) =.) ( ; ) and (; ).) ( ; ) and (; ).) a) = + b) + = ( ) or =.) a) ( ; ) and ( ; ) b) ( ; ) and (; ) Practice Problems - Answers.) a) ( ) + = 9 b) + ( + ) = c) ( + 8) + ( ) =.) ( ) + ( + ) = center: (; ) radius: ) a) ( ; ) and ( ; ) b) ( 7; ) c) there is no such point d) ; p and ; + p.) a) ( ; 9) b) (; ) and (9; ) c) the don t intersect d) ( ; ) and (; ).) a) ( + 8) = or = + b) 7 ( ) = + or = 7 7.) a) (; ) b) (7; ) and (; ) c) the circles do not intersect c copright Hidegkuti, Powell, 8 Last revised: March 9,

3 Lecture Notes Circles - Part page Sample Problems - Solutions. Find an equation for the circle centered at (; ) with radius r = units. Solution: Let P (; ) be a general point on the circle. If the point P is on the circle, then it must be units awa from the point C (; ) : Consider the right triangle shown on the picture below. The sides of this triangle are j j, j + j, and : The Phtagorean theorem, stated for this triangle is ( ) + ( + ) = ; which is the equation for the circle.. Graph the equation + + = ( ) Solution: Complete the squares to read the center and radius. + + Thus the center is ( ; ) and the radius : + + = ( ) + + = + + = = ( + ) + ( ) = ( + ) + ( ) = c copright Hidegkuti, Powell, 8 Last revised: March 9,

4 Lecture Notes Circles - Part page. Consider the circle ( ) + ( + ) =. Find all points on the circle with coordinate being. Solution: We set = and solve for in the circle s equation. Because the equation is quadratic, we ma obtain two or one or no solution. It is also helpful to sketch a graph of the circle; it is centered at (; ) and has a radius of p. ( ) + ( + ) = ( ) = ( ) + ( ) = ( + ) ( ) = ( ) + = ( + ) ( ) = ( ) = = = Therefore, there are two points on this circle with coordinate ; and the are ( ; ) and (; ) Find the coordinates of all points where the circle + ( ) = and the line = + 8 Solution: We need to solve the following sstem of equations: + ( ) = = + 8 We will use substitution. We substitute = + 8 in the rst equation and solve for. + ( + 8 ) = = + ( + ) = = + + = ( ) ( + ) = + = = = We now use the second equation to nd the value belonging to the values we just obtained. If =, then = () + 8 = : If =, then = ( ) + 8 = : Thus the two points are (; ) and ( ; ). We check: both points should be on both the circle and the line. c copright Hidegkuti, Powell, 8 Last revised: March 9,

5 Lecture Notes Circles - Part page. Consider the circle + =. Find an equation for the tangent line drawn to the circle at the point (; ). Solution: The tangent line is perpendicular to the radius drawn to the point of tangenc. We can easil nd the slope of this radius as the slope of the line segment connecting the center of the circle, (; ) and the point of tangenc, (; ). We use the slope formula, slope = m = rise run = = = Since perpendicular to a line with slope, the tangent line must have slope, the negative reciprocal of. It must also pass through the point (; ) : The point-slope form of this line s equation is then = ( ) : We simplif this and obtain = b) Consider the circle + + =. Find an equation for the tangent line drawn to the circle at the point (; ). Solution: We rst transform the equation of the circle to determine its center s coordinates = + + = = ( + ) 9 + ( ) = ( + ) + ( ) = Thus the center of the circle is ( ; ) and its radius is c copright Hidegkuti, Powell, 8 Last revised: March 9,

6 Lecture Notes Circles - Part page The tangent line is perpendicular to the radius drawn to the point of tangenc. We can easil nd the slope of this radius: the slope of the line segment connecting the center ( ; ) and the point of tangenc (; ) is slope = m = rise run = = ( ) = Since perpendicular to a line with slope, the tangent line must have slope, the negative reciprocal of. The tangent line must also pass through the point (; ) : The point-slope form of this line s equation is then + = ( ). We can simplif this and obtain the slope-intercept form, = a) Find the points where the circles ( + ) + ( + ) = and ( + ) + ( ) = Solution: We need to solve the following sstem: ( + ) + ( + ) = ( + ) + ( ) = We multipl out the complete squares and combine like terms in both equations = = = = = = c copright Hidegkuti, Powell, 8 Last revised: March 9,

7 Lecture Notes Circles - Part page 7 We will multipl the second equation b and add the two equations. This will cancel out all quadratic terms. The sum of the two equations is = 7 + = + = divide both sides b + = = We substitute this into the rst equation and solve ( + ) + ( + ) = and = ( + ) + ( + ) = + A + ( + ) = ( + ) + ( + ) = = + = = ( ) = =) = = We now nd the values belonging to the values, using =. If =, then = = and if =, then = =. Thus the two circles intersect at the points ( ; ) and ( ; ) b) Find the points where the circles ( + ) + ( + ) = and ( ) + ( ) = Solution: We need to solve the following sstem: ( + ) + ( + ) = ( ) + ( ) = c copright Hidegkuti, Powell, 8 Last revised: March 9,

8 Lecture Notes Circles - Part page 8 We multipl out the complete squares and combine like terms in both equations = = = + = We will multipl the second equation b and add the two equations. This will cancel out all quadratic terms. The sum of the two equations is = + + = 8 + = divide both sides b + = We now solve for + = = + We substitute this into the rst equation and solve for. ( + ) + ( + ) = and = + ( + ) C {z A } = + 7 ( + ) + = ( + ) ( + 7) + = since + = 7 a b = a ( + ) ( + 7) + = multipl b ( + ) + ( + 7) = = = = = ( ) ( + ) = =) = = We now nd the values belonging to the values, using = () + = = and if =, then = intersect at the points ( ; ) and (; ) : ( ) + b +. If = ; then =. Thus the two circles For more documents like this, visit our page at and click on Lecture Notes. questions or comments to [email protected]. c copright Hidegkuti, Powell, 8 Last revised: March 9,