hsn.uk.net Higher Mathematics UNIT OUTCOME 4 Circles Contents Circles 119 1 Representing a Circle 119 Testing a Point 10 3 The General Equation of a Circle 10 4 Intersection of a Line an a Circle 1 5 Tangents to Circles 13 6 Equations of Tangents to Circles 14 7 Intersection of Circles 16 HSN400 This ocument was prouce specially for the HSN.uk.net website, an we require that any copies or erivative works attribute the work to Higher Still Notes. For more etails about the copyright on these notes, please see http://creativecommons.org/licenses/by-nc-sa/.5/scotlan/
Unit Circles OUTCOME 4 Circles 1 Representing a Circle The equation of a circle with centre ( a, b ) an raius r units is ( a) + ( y b) r. This is given in the eam. For eample, the circle with centre (, 1) an raius 4 units has equation: ( ) ( y ) ( ) ( y ) + + 1 4 + + 1 16. Note that the equation of a circle with centre ( 0, 0 ) is of the form + y r, where r is the raius of the circle. EXAMPLES 1. Fin the equation of the circle with centre ( 1, 3) an raius 3 units. a + y b r ( 1) + y ( 3) 3 ( ) ( y ) 1 + + 3 3.. A is the point ( 3,1) an B( 5,3 ). Fin the equation of the circle which has AB as a iameter. The centre of the circle is the mipoint of AB; 5 3 3 + 1 C mipoint, ( 1, ). AB The raius r is the istance between A an C: r ( 1 ( 3) ) + ( 1) 4 + 1 17. So the equation of the circle is ( ) ( y ) 1 + 17. Note You coul also use the istance between B an C, or half the istance between A an B. Page 119 HSN400
Unit Circles Testing a Point Given a circle with centre ( a, b ) an raius r units, we can etermine whether a point ( p, q ) lies within, outwith or on the circumference using the following rules: p a + q b < r the point lies within the circle p a + q b r the point lies on the circumference of the circle p a + q b > r the point lies outwith the circle. EXAMPLE + y + 5 9. A circle has the equation Determine whether the points (,1 ), ( 7, 3) an ( 3, 4) lie within, outwith or on the circumference of the circle. Point (,1 ) : ( ) + ( y + 3) ( ) + ( 1+ 5) 0 + 6 36 > 9 So outwith the circle. Point ( 7, 3) : ( ) + ( y + 3) ( 7 ) + ( 3 + 5) 5 + 9 So on the circumference. Point ( 3, 4) : ( ) + ( y + 3) ( 3 ) ( 4 5) + + 1 + 1 < 9 So within the circle. 3 The General Equation of a Circle The equation of any circle can be written in the form where the centre is ( g, f ) This is given in the eam. + + + + 0 y g fy c an the raius is g + f c units. Note that the above equation only represents a circle if since: + > 0, g f c if g + f c < 0 then we cannot obtain a real value for the raius, since we woul have to square root a negative; if g + f c 0 then the raius is zero the equation represents a point rather than a circle. Page 10 HSN400
Unit Circles EXAMPLES 1. Fin the raius an centre of the circle with equation + y + 4 8y + 7 0. Comparing with g 4 so g f 8 so f 4 c 7 + + + + 0: y g fy c Centre is, ( g f ) (, 4) Raius is. Fin the raius an centre of the circle with equation + y 6 + 10 y 0. The equation must be in the form each term by : + 3 + 5 1 0 y y Now compare with g 3 so g 3 f 5 so f 5 c 1 3. Eplain why y g fy c g + f c ( ) 4 7 + 4 + 16 7 y g fy c + + + + 0: Centre is, y y ( g f ) 3 5 (, ) 13 units. + + + + 0, so ivie Raius is g + f c 3 5 + + 1 9 5 4 4 + 4 + 4 38 4 38 units. + + 4 8 + 9 0 is not the equation of a circle. Comparing with + y + g + fy + c 0 : g 4 so g g + f c + ( 4) 9 f 8 so f 4 9 < 0. c 9 The equation oes not represent a circle since satisfie. + > 0 is not g f c Page 11 HSN400
Unit Circles 4. For which values of k oes a circle? Comparing with y k y k k y g fy c c k k + 4 + + 4 0 represent + + + + 0 : g k so g k To represent a circle, we nee f 4 so f + 4. 4 Intersection of a Line an a Circle g f c ( ) k k k + > 0 + 4 + 4 > 0 k + 8 > 0 k < 8. A straight line an circle can have two, one or no points of intersection: two intersections one intersection no intersections If a line an a circle only touch at one point, then the line is a tangent to the circle at that point. To fin out how many times a line an circle meet, we can use substitution. EXAMPLES 1. Fin the points where the line with equation y 3 intersects the circle with equation + y 0. + y 0 + ( 3 ) 0 + 9 0 10 0 ± y 3( ) 3 y 3( ) 3. So the circle an the line meet at (, 3 ) an (, 3 ). Remember m m m ( ab) a b. Page 1 HSN400
Unit Circles. Fin the points where the line with equation y + 6 an circle with equation + y + + y 8 0 intersect. Substitute y + 6 into the equation of the circle: + ( + 6) + + ( + 6) 8 0 + ( + 6)( + 6) + + 4 + 1 8 0 + 4 + 4 + 36 + + 4 + 1 8 0 5 + 30 + 40 0 ( ) 5 + 6 + 8 0 ( )( ) + + 4 0 + 0 + 4 0 4 y ( ) + 6 y ( 4) + 6. So the line an circle meet at (, ) an ( 4, ). 5 Tangents to Circles As we have seen, a line is a tangent if it intersects the circle at only one point. To show that a line is a tangent to a circle, the equation of the line can be substitute into the equation of the circle, an solve there shoul only be one solution. EXAMPLE Show that the line with equation + y 4 is a tangent to the circle with equation + y + 6 + y 0. Substitute y using the equation of the straight line: y y + + 6 + 0 + 4 + 6 + 4 0 ( ) + 4 4 + 6 + 4 0 + 16 8 + + 6 + 8 0 4 + 0 ( ) + 1 0 + 1 0 Page 13 HSN400
Unit Circles Then (i) factorise or (ii) use the iscriminant + 1 0 ( 1)( 1) 0 1 0 1 1 0 1. Since the solutions are equal, the line is a tangent to the circle. a 1 b c 1 + 1 0 b 4ac ( ) 4( 1 1) 4 4 0. Since b 4ac 0, the line is a tangent to the circle. Note If the point of contact is require then metho (i) is more efficient. To fin the point, substitute the value foun for into the equation of the line (or circle) to calculate the corresponing y-coorinate. 6 Equations of Tangents to Circles If the point of contact between a circle an a tangent is known, then the equation of the tangent can be calculate. If a line is a tangent to a circle, then a raius will meet the tangent at right angles. The graient of this raius can be calculate, since the centre an point of contact are known. Using mraius mtangent 1, the graient of the tangent can be foun. The equation can then be foun using y b m( a), since the point is known, an the graient has just been calculate. Page 14 HSN400
Unit Circles EXAMPLE Show that A ( 1, 3 ) lies on the circle equation of the tangent at A. Substitute point into equation of circle: + y + 6 + y 1 + 3 + 6( 1) + ( 3) 1+ 9 + 6 + 6 + + 6 + 0 an fin the y y 0. Since this satisfies the equation of the circle, the point must lie on the circle. Fin the graient of the raius from ( 3, 1) to ( 1, 3 ): y y1 mraius 1 3 + 1 1 + 3 1. So m 1 since m m 1. tangent raius tangent Fin equation of tangent using point ( 1, 3 ) an graient m 1: y b m( a) y 3 ( 1) y 3 + 1 y + 4 + y 4 0. Therefore the equation of the tangent to the circle at A is + y 4 0. Page 15 HSN400
Unit Circles 7 Intersection of Circles Consier two circles with raii r 1 an r with r > r. 1 Let be the istance between the centres of the two circles. r 1 r > r1 + r The circles o not touch. r1 + r r1 r < < r1 + r The circles touch eternally. The circles meet at two istinct points. Note Don t try to memorise this, just try to unerstan why each one is true. r1 r The circles touch internally. < r1 r The circles o not touch. EXAMPLES 1. Circle P has centre ( 4, 1) y y an raius units, circle Q has equation + + 6 + 1 0. Show that the circles P an Q o not touch. To fin the centre an raius of Q: Compare with + y + g + fy + c 0: g so g 1 Centre is ( g, f ) Raius rq g + f c f 6 so f 3 ( 1, 3 ). 1+ 9 1 c 1. 9 3 units. Page 16 HSN400
Unit Circles We know P has centre ( 4, 1) an raius r P units. So the istance between the centres ( 1+ 4) + ( 3 + 1) 5 + ( ) 9 5. 39 units (to.p.). Since r P + r Q 3 + 5 <, the circles P an Q o not touch.. Circle R has equation equation ( ) ( y ) eternally. + 4 4 0, an circle S has y y 4 + 6 4. Show that the circles R an S touch To fin the centre an raius of R: Compare with g so g 1 f 4 so f + + + + 0: y g fy c c 4. Centre is, To fin the centre an raius of S: compare with r a 4 b 6 4 so r. ( g f ) ( 1, ). a y b r +. Centre is, ( a b) ( ) 4, 6. Raius rr g + f c S ( 1) + ( ) + 4 9 3 units. Raius r units. So the istance between the centres ( 1 4) + ( 6) ( 3) ( 4) + 5 5 units. Since r R + r S 3 + 5, the circles R an S touch eternally. Page 17 HSN400