# Coordinate Geometry Locus II

## Presentation on theme: "Coordinate Geometry Locus II"— Presentation transcript:

Coordinate Geometry Locus II
By Mr Porter

Definition 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. The standard circle has center at the origin, (0, 0) and radius, r. It LOCUS (equation) is of the form x2 + y2 = r2 (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 = r2 y (h,k) r x

Derive the Locus of a Circle.
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] The distance AP = r units. Draw a small diagram. y A(h,k) r x P(x,y) Setting the distance formula to equal r: Substitute and square both sides This is the best form, but you are required to expand. Since we are talking distance on the coordinate number plane, we use the distance formula: Rearrange. The locus of P is x2 + y2 – 2hx – 2ky + h2 + k2 – r2 = 0. Note: The distance formula is used for the majority of questions relating to locus. You do NOT need to know PROOF, just the method!

Derive the Locus of a Circle.
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). The distance AP = 4 units. Draw a small diagram. y A(1,-3) d =4 x P(x,y) Setting the distance formula to equal 4: Substitute and square both sides This is the best form, but you are required to expand. Since we are talking distance on the coordinate number plane, we use the distance formula: Rearrange. The locus of P is x2 + y2 – 2x + 6y – 6 = 0. Note: The distance formula is used for the majority of questions relating to locus. Note: This is the DERIVE method!

Find the equation of the circle with centre (4,-5) and radius 8 units.
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) 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 = r2 Make the substitutions. (x – 4)2 + (y – -5)2 = 82 (x – 4)2 + (y +5)2 = 64 This is the best form, but you are required to expand. x2 – 8x y2 + 10y +25 = 64 Rearrange. x2 + y2 – 8x + 10y – 23 = 0 We could DERIVE the equation of the Locus But, since we know it’s a CIRCLE, we can use the general formula. The locus of P is x2 + y2 – 8x + 10y – 23 = 0. The general circle has center at (h, k) and radius, r. It LOCUS (equation) is of the form (x – h)2 + (y – k)2 = r2 Note: This is the FIND method!

Data: Centre (h,k) = (-2, 5) and radius r = ?
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) Data: Centre (h,k) = (-2, 5) and radius r = ? Let P(3, -1) lie on the circle. Radius, r: Substitute the 2 points values. Evaluate under √ using calculator. Make the substitutions. {Note: r2 = 61 } (x – -2)2 + (y – 5)2 = 61 So, (x – h)2 + (y – k)2 = r2 Since we know it’s a CIRCLE, we can use the general formula. But the radius is unknown? x2 + 4x y2 – 10y +25 = 64 This is the best form, but you are required to expand. (x + 2)2 + (y – 5)2 = 61 We must use the distance formula to calculate the radius of the circle. Rearrange. x2 + y2 + 4x – 10y – 32 = 0 The general circle has center at (h, k) and radius, r. It LOCUS (equation) is of the form (x – h)2 + (y – k)2 = r2 The locus of P is x2 + y2 + 4x – 10y – 32 = 0.

This is of the form: (x – h)2 + (y – k)2 = r2.
Reverse Type Question. Example 5 Given the equation of a circle is, x2 + y2 – 4x + 6y – 12 = 0, find it’s centre (h, k) and length of it’s radius, r. Rearrange, algebra on LHS and numbers on RHS. (x’s together, y’s together!) In this question, you need to re-arrange the given equation in the form (x – h)2 + (y – k)2 = r2, centre (h, k) and radius r. x2 – 4x + y2 + 6y = 12 To complete the square for x, take (-4) ÷ 2 and square = [(-4)÷2] 2 and add to both sides. Here, we need to complete the square method of FACTORISATION for both x and y, simultaneously. To complete the square for y, take (+6) ÷ 2 and square = [(+6)÷2] 2 and add to both sides. x2 – 4x y2 + 6y + 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 = r2. centre (h, k) = (2, -3), radius r = 5 .

This is of the form: (x – h)2 + (y – k)2 = r2.
Example 6 Find the centre and radius of the circle with the equation, x2 + y2 +6x – 2y + 3 = 0. Rearrange, algebra on LHS and numbers on RHS. (x’s together, y’s together!) In this question, you need to re-arrange the given equation in the form (x – h)2 + (y – k)2 = r2, centre (h, k) and radius r. x2 + 6x + y2 – 2y = -3 To complete the square for x, take (+6) ÷ 2 and square = [(+6)÷2] 2 and add to both sides. Here, we need to complete the square method of FACTORISATION for both x and y, simultaneously. To complete the square for y, take (-2) ÷ 2 and square = [(-2)÷2] 2 and add to both sides. x2 +6x y2 – 2y + 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 = r2. centre (h, k) = (-3, 1), radius r =