In the diagram, BE and CD are diameters of a circle having M as centre - NSC Mathematics - Question 10 - 2021 - Paper 2

Since AE is perpendicular to CD and intersects it at F, we have a right angle. Therefore, the angle at the circle's center M connects with the angle at point A and we can express it as:

$∠M1 = 180° - 2x$

This is because the angle at the center is twice that of the angle at the circumference.

To prove that AD is a tangent to the circle at point A, we note that:

• Chord CD is perpendicular to the radius at point F (the point of intersection) and that the angle ∠C is given by the chord. Therefore, by the converse of the tangent-chord theorem, since AE ⊥ CD, then AD is indeed tangent at A.

Given that AB is a diameter of the circle and equals 24 units, we can find the radius as:

r = rac{AB}{2} = rac{24}{2} = 12 ext{ units}

Using the Pythagorean theorem in triangle ACF:

$AC^2 = AF^2 + CF^2$ $AC^2 = AE^2 + 6^2$

Since AC also equals the radius:

$12^2 = AE^2 + 36$

Thus:

$144 = AE^2 + 36$ $AE^2 = 108$ $AE = ext{√108} = 12 ext{√3} ext{ or approximately } 20.78 ext{ units}$