1. Coordinate Geometry Locus III - Problems By Mr Porter

2. Example 1 For the circle with equation x 2 + y 2 + 6x – 8y = 0, find the coordinates of the centre and the length of the radius. We need to write the equation of the circle in the general form (x – h) 2 + (y – k) 2 = r 2 Therefore, you need to complete the square for both ‘x’ and ‘y’ simultaneously, while writing all the numerical values on the RHS. x 2 + y 2 + 6x – 8y = 0 Group “x’s” and “y’s” separately on the LHS. x 2 + 6x + y 2 – 8y = 0 To complete the square, use the coefficient of ‘x’, +6x  +6 { (+ 6 / 2 ) 2 = 9. Add ‘9’ to both sides.} x 2 + 6x + 9 + y 2 – 8y = 0 + 9 To complete the square, use the coefficient of ‘y’, -8x  -8 { (- 8 / 2 ) 2 = 16. Add ‘16’ to both sides.} x 2 + 6x + 9 + y 2 – 8y + 16 = 0 + 9 + 16 Now, complete the square for ‘x’ and ‘y’ on LHS by FACTORISATION, clean up RHS. (x + 3) 2 + (y – 4) 2 = 25 This is of the form: (x – h) 2 + (y – k) 2 = r 2 centre (h, k) and radius = r Hence, Centre (h, k) = (-3, 4) and radius r = 5

3. Example 2 The equation of the circle is x 2 + y 2 + 4x – 2y – 20 = 0. Find the length of the tangent to this circle from the point (5, 2). We need to write the equation of the circle in the general form (x – h) 2 + (y – k) 2 = r 2 Therefore, you need to complete the square for both ‘x’ and ‘y’ simultaneously, while writing all the numerical values on the RHS. x 2 + 4x + y 2 – 2y = 0 + 20 Group “x’s” and “y’s” separately on the LHS, move numbers to RHS. To complete the square, use the coefficient of ‘x’, +4x  +4 { (+ 4 / 2 ) 2 = 4. Add ‘4’ to both sides. Do the same for ‘y’.} x 2 + 4x + 4 + y 2 – 2y + 1 = 20 + 4 + 1 Now, complete the square for ‘x’ and ‘y’ on LHS by FACTORISATION, clean up RHS. (x + 2) 2 + (y – 1) 2 = 25 This is of the form: (x – h) 2 + (y – k) 2 = r 2 centre (h, k) and radius = r Hence, Centre (h, k) = (-2, 1) and radius r = 5 We need to construct a simple diagram to help solve the problem; “Length of the tangent = d.” Fact: The tangent (AC) from an external point (A) is perpendicular to the radius (r = OC) at the point of contact (C). The relationship for r, h and d is PYTHAGORS’ Theorem: h 2 = r 2 + d 2 We need to find ‘d’. We already have r = 5, but the hypotenuse is the distance AO {distance between two points}.

4. (Example 2 continued) We need to find ‘d’. We already have r = 5, but the hypotenuse is the distance AO {distance between two points (-2, 1) and (5, 2)}. Length of hypotenuse, h = distance AO. Length of d = distance AC. The length of the tangent is d = 5 units. The length of the second tangent AD = AC.

5. Example 3 Find the locus of a point P(x, y) which moves in the coordinate number plane such that PA = 2PB, where A= (-2, 4) and B(4, 1). Show that it is a circle. A little diagram to help! A(-2,4) P(x,y) B(4,1) Given: distance AP = 2 x distance PB Substitute values and square both side. [Don’t forget to square the ‘2’!] Expand both side. Take care with the RHS! Rearrange. To show that it is a circle, we need to find it’s centre and radius. We must write it in the general form (x – h) 2 + (y – k) 2 = r 2 Rearrange and complete the square for x and y. This is in the form of the locus of a circle: (x – h) 2 + (y – k) 2 = r 2, centre (h, k) radius r. Centre (6, 0) and radius r units.

6. Example 4 Find the locus of point P(x, y), such that its position from A(1,2) and B(4,-1) is in the ratio AP : PB is 1 : 2. Describe the locus. What is the meaning of AP : PB = 1 : 2? The length of AP fits 2 times into length PB. Therefore 2 x AP = PB! AP: PB 1: 2now, cross multiply the ratios. 2 AP= 1 PB, apply the distance formula. Substitute values and square both side. [Don’t forget to square the ‘2’!] 4[(x – 1) 2 + (y – 2) 2 ] = (x – 4) 2 +(y + 1) 2 4[x 2 -2x + 1 + y 2 – 4y + 4] = x 2 – 8x + 16 + y 2 + 2y +1 4x 2 -8x + 4 + 4y 2 – 16y + 16 = x 2 – 8x + 16 + y 2 + 2y +1 3x 2 + 3y 2 – 18y = -30 Divide by 3 Complete the square for x and y. x 2 + y 2 – 6y = -5 x 2 + y 2 – 6y + 9 = -5 + 9 (x – 0) 2 + (y – 3) 2 = 4 This is in the form of a circle: (x – h) 2 + (y – k) 2 = r 2, centre (h, k) radius r. Centre (0, 3) and radius, r = 2 units. The locus is x 2 + y 2 – 6y + 5 = 0.

7. Example 5 A(-1, 3) and B(3, 1) are two point on the coordinate number plane. Find the locus of the point P(x, y) such that PA | PB. Since PA | PB, the product of the gradients is -1 i.e. m AP x m BP = -1 Then, (y – 3)( y – 1) = -1 (x +1) (x – 3) expand, y 2 – 4y + 3 = -1 [x 2 – 2x – 3] y 2 – 4y + 3 = -x 2 + 2x + 3 x 2 + y 2 – 2x – 4y = 0, the locus of P Rearrange and complete the square for x and y. x 2 – 2x + y 2 – 4y = 0 x 2 – 2x + 1+ y 2 – 4y + 4 = 0 + 1 + 4 (x – 1) 2 + (y – 2) 2 = 5 This is in the form of a circle: (x – h) 2 + (y – k) 2 = r 2, centre (h, k) radius r. Centre (1, 2) and radius, r = units.