Coordinate Geometry Locus III - Problems By Mr Porter.

Coordinate GeometryLocus III - Problems

By Mr Porter

Example 1For the circle with equation x2 + y2 + 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 = r2

Therefore, you need to complete the square for both ‘x’ and ‘y’ simultaneously, while writing all the numerical values on the RHS.

x2 + y2 + 6x – 8y = 0Group “x’s” and “y’s” separately on the LHS.

x2 + 6x + y2 – 8y = 0To complete the square, use the coefficient of ‘x’, +6x +6{ (+6/2)2 = 9. Add ‘9’ to both sides.}

x2 + 6x + 9 + y2 – 8y = 0 + 9To complete the square, use the coefficient of ‘y’, -8x -8{ (-8/2)2 = 16. Add ‘16’ to both sides.}

x2 + 6x + 9 + y2 – 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 = r2

centre (h, k) and radius = r

Hence,Centre (h, k) = (-3, 4) and radius r = 5

Example 2

The equation of the circle is x2 + y2 + 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 = r2

Therefore, you need to complete the square for both ‘x’ and ‘y’ simultaneously, while writing all the numerical values on the RHS.

x2 + 4x + y2 – 2y = 0 + 20Group “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’.}

x2 + 4x + 4 + y2 – 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 = r2

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:h2 = r2 + d2

We need to find ‘d’. We already have r = 5, but the hypotenuse is the distance AO {distance between two points}.

(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.

Example 3Find 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 = r2

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 = r2 , centre (h, k) radius r.

Centre (6, 0) and radius r units.

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 : 2 now, 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[x2 -2x + 1 + y2 – 4y + 4] = x2 – 8x + 16 + y2 + 2y +1

4x2 -8x + 4 + 4y2 – 16y + 16 = x2 – 8x + 16 + y2 + 2y +1

3x2 + 3y2 – 18y = -30

3x2 + 3y2 – 18y = -30 Divide by 3

Complete the square for x and y.x2 + y2 – 6y = -5

x2 + y2 – 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 = r2 , centre (h, k) radius r.Centre (0, 3) and radius, r = 2 units.

The locus is x2 + y2 – 6y + 5 = 0 .

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. mAP x mBP = -1

Then, (y – 3)( y – 1) = -1 (x +1) (x – 3)

expand, y2 – 4y + 3 = -1 [x2 – 2x – 3]

y2 – 4y + 3 = -x2 + 2x + 3

x2 + y2 – 2x – 4y = 0 , the locus of P

Rearrange and complete the square for x and y.

x2 – 2x + y2 – 4y = 0

x2 – 2x + 1+ y2 – 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 = r2 , centre (h, k) radius r.Centre (1, 2) and radius, r = units.