O(0 ; 0) and N(2 ; y) are points on the circumference of a circle with centre M(4 ; 2). Determine the equation of the circle.

1. (x − 4)² + (y − 2)² = 20
2. (x − 4)² + (y − 2)² = 4
3. (x + 4)² + (y + 2)² = 20
4. x² + y² = 20

Answer: (x − 4)² + (y − 2)² = 20

Find the radius using the distance from M(4 ; 2) to O(0 ; 0):

r = √[(4 − 0)² + (2 − 0)²] = √[16 + 4] = √20

So r² = 20

* The equation uses (x − a)² + (y − b)² = r² where (a ; b) is the centre.

N(2 ; y) lies on the circle with equation (x − 4)² + (y − 2)² = 20, and y > 0. Calculate the value of y.

1. y = 6
2. y = 2
3. y = −2
4. y = 4

Answer: y = 6

Substitute N(2 ; y) into the circle's equation:

(2 − 4)² + (y − 2)² = 20

(−2)² + (y − 2)² = 20

4 + (y − 2)² = 20

(y − 2)² = 16

y − 2 = ±4

y = 6 or y = −2

* Both solutions satisfy the equation, but the diagram shows N in the upper half-plane.

The tangent to the circle at O(0 ; 0) meets the tangent at N(2 ; 6) at point R. Determine the equation of the tangent OR.

1. y = 2x
2. y = −2x
3. y = 12x
4. y = −12x

Answer: y = −2x

The radius MO goes from M(4 ; 2) to O(0 ; 0).

Gradient of MO = (2 − 0) ÷ (4 − 0) = ½

The tangent at O is ⊥ to the radius MO.

Gradient of tangent OR = −1 ÷ (½) = −2

Tangent passes through O(0 ; 0), so c = 0.

* A tangent is always perpendicular to the radius at the point of tangency.

The tangent at N(2 ; 6) and the tangent OR ( y = −2x ) meet at R. Calculate the coordinates of R.

1. R(−1 ; 2)
2. R(1 ; −2)
3. R(−2 ; 4)
4. R(0 ; 0)

Answer: R(−2 ; 4)

Step 1 — Find the equation of the tangent at N(2 ; 6):

Gradient of radius MN: M(4 ; 2) → N(2 ; 6): m = (6 − 2) ÷ (2 − 4) = 4 ÷ (−2) = −2

The tangent at N is ⊥ to MN, so its gradient = −1 ÷ (−2) = ½

Tangent at N: y − 6 = ½(x − 2) ⟹ y = ½x + 5

Step 2 — Solve simultaneously with tangent OR ( y = −2x ):

−2x = ½x + 5

−2x − ½x = 5 ⟹ −52x = 5

x = −2, and y = −2(−2) = 4

* Since both tangents are drawn from the external point R, the tangent lengths OR and NR are equal.

Determine, with a reason, the type of quadrilateral represented by MNRO.

1. Kite — MO = MN (equal radii) and OR = NR (equal tangents from R)
2. Rectangle — all four angles are 90°
3. Rhombus — all four sides are equal
4. Parallelogram — both pairs of opposite sides are parallel

Answer: Kite — MO = MN (equal radii) and OR = NR (equal tangents from R)

MO and MN are both radii, so MO = MN — one pair of adjacent sides is equal.

OR and NR are tangents to the circle from the same external point R, so OR = NR — the other pair of adjacent sides is equal.

A quadrilateral with two pairs of equal adjacent sides is a kite.

* It is not a rectangle/rhombus/square: although ∠MOR = ∠MNR = 90° (radius ⊥ tangent), MO ≠ OR in general, so the sides are not all equal and the figure is not a parallelogram.