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Coordinate Geometry Locus II By Mr Porter. A circle may be defined as the set of all points, P, in a plane at a given distance from a fixed point in the.

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Presentation on theme: "Coordinate Geometry Locus II By Mr Porter. A circle may be defined as the set of all points, P, in a plane at a given distance from a fixed point in the."— Presentation transcript:

1 Coordinate Geometry Locus II By Mr Porter

2 A circle may be defined as the set of all points, P, in a plane at a given distance from a fixed point in the plane. The fixed point is the centre of the circle and the fixed distance is the radius. Definition The standard circle has center at the origin, (0, 0) and radius, r. It LOCUS (equation) is of the form x 2 + y 2 = r 2 (0,0) r x y The general circle has center at (h, k) and radius, r. It LOCUS (equation) is of the form (x – h) 2 + (y – k) 2 = r 2 y (h,k) r x

3 Example 1: Find the locus of point, P(x,y) such that it remains a fixed distance r, from a fixed point, (h,k). [ This is the general equation of a circle] Derive the Locus of a Circle. Since we are talking distance on the coordinate number plane, we use the distance formula: Note: The distance formula is used for the majority of questions relating to locus. Draw a small diagram. y A(h,k) r x P(x,y) The distance AP = r units. Setting the distance formula to equal r: Substitute and square both sides This is the best form, but you are required to expand. Rearrange. The locus of P is x 2 + y 2 – 2hx – 2ky + h 2 + k 2 – r 2 = 0. You do NOT need to know PROOF, just the method!

4 Example 2: Find the locus of point, P(x,y) such that it remains a fixed distance of 4 units, from a fixed point, A(1,-3). Derive the Locus of a Circle. Since we are talking distance on the coordinate number plane, we use the distance formula: Note: The distance formula is used for the majority of questions relating to locus. Draw a small diagram. y A(1,-3) d =4 x P(x,y) The distance AP = 4 units. Setting the distance formula to equal 4: Substitute and square both sides This is the best form, but you are required to expand. Rearrange. The locus of P is x 2 + y 2 – 2x + 6y – 6 = 0. Note: This is the DERIVE method!

5 Example 3: Find the equation of the circle with centre (4,-5) and radius 8 units. Draw a small diagram. y A(4,- 5) d = 8 x P(x,y) We could DERIVE the equation of the Locus But, since we know it’s a CIRCLE, we can use the general formula. The general circle has center at (h, k) and radius, r. It LOCUS (equation) is of the form (x – h) 2 + (y – k) 2 = r 2 Data: Centre (h,k) = (4,-5) and radius r = 8 Let P(x, y) lie on the circle. So, (x – h) 2 + (y – k) 2 = r 2 Make the substitutions. (x – 4) 2 + (y – -5) 2 = 8 2 (x – 4) 2 + (y +5) 2 = 64 This is the best form, but you are required to expand. x 2 – 8x + 16 + y 2 + 10y +25 = 64 Rearrange. x 2 + y 2 – 8x + 10y – 23 = 0 The locus of P is x 2 + y 2 – 8x + 10y – 23 = 0. Note: This is the FIND method!

6 Example 4: Find the equation of the circle with centre (-2, 5) and passing through the point P(3,-1). Draw a small diagram. y A(-2,5) d = r x P(3, -1) The general circle has center at (h, k) and radius, r. It LOCUS (equation) is of the form (x – h) 2 + (y – k) 2 = r 2 Data: Centre (h,k) = (-2, 5) and radius r = ? Let P(3, -1) lie on the circle. Since we know it’s a CIRCLE, we can use the general formula. But the radius is unknown? We must use the distance formula to calculate the radius of the circle. Radius, r: Substitute the 2 points values. Evaluate under √ using calculator. So, (x – h) 2 + (y – k) 2 = r 2 Make the substitutions. {Note: r 2 = 61 } (x – -2) 2 + (y – 5) 2 = 61 (x + 2) 2 + (y – 5) 2 = 61 Rearrange. x 2 + y 2 + 4x – 10y – 32 = 0 The locus of P is x 2 + y 2 + 4x – 10y – 32 = 0. x 2 + 4x + 4 + y 2 – 10y +25 = 64 This is the best form, but you are required to expand.

7 Reverse Type Question. Example 5 Given the equation of a circle is, x 2 + y 2 – 4x + 6y – 12 = 0, find it’s centre (h, k) and length of it’s radius, r. In this question, you need to re-arrange the given equation in the form (x – h) 2 + (y – k) 2 = r 2, centre (h, k) and radius r. Here, we need to complete the square method of FACTORISATION for both x and y, simultaneously. Rearrange, algebra on LHS and numbers on RHS. (x’s together, y’s together!) x 2 – 4x + y 2 + 6y = 12 To complete the square for x, take (-4) ÷ 2 and square = [(-4)÷2] 2 and add to both sides. To complete the square for y, take (+6) ÷ 2 and square = [(+6)÷2] 2 and add to both sides. x 2 – 4x + 4 + y 2 + 6y + 9 = 12 + 4 + 9 Factorise for x and y and evaluate RHS. (x – 2) 2 + (y +3 ) 2 = 25 This is of the form: (x – h) 2 + (y – k) 2 = r 2. centre (h, k) = (2, -3), radius r = 5.

8 Example 6 Find the centre and radius of the circle with the equation, x 2 + y 2 +6x – 2y + 3 = 0. In this question, you need to re-arrange the given equation in the form (x – h) 2 + (y – k) 2 = r 2, centre (h, k) and radius r. Here, we need to complete the square method of FACTORISATION for both x and y, simultaneously. Rearrange, algebra on LHS and numbers on RHS. (x’s together, y’s together!) x 2 + 6x + y 2 – 2y = -3 To complete the square for x, take (+6) ÷ 2 and square = [(+6)÷2] 2 and add to both sides. To complete the square for y, take (-2) ÷ 2 and square = [(-2)÷2] 2 and add to both sides. x 2 +6x + 9 + y 2 – 2y + 1 = -3 + 9 + 1 Factorise for x and y and evaluate RHS. (x + 3) 2 + (y – 1 ) 2 = 7 This is of the form: (x – h) 2 + (y – k) 2 = r 2. centre (h, k) = (-3, 1), radius r =.


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