Circle Geometry & Proof

The angle in a semicircle is a right angle: an angle subtended by a diameter at the circumference is always $90^\circ$, wherever $C$ sits on the arc.

The angle at the centre is twice the angle at the circumference standing on the same arc, so $\angle ACB = \tfrac{1}{2}\angle AOB = \tfrac{1}{2}(80^\circ) = 40^\circ$. (160° doubled it the wrong way.)

Opposite angles of a cyclic quadrilateral are supplementary (add to $180^\circ$): $\angle C = 180^\circ - 95^\circ = 85^\circ$. (Equal angles would only happen if both were $90^\circ$.)

Use the first chord to find the radius: half of it is $8$ cm, so $r = \sqrt{8^2 + 6^2} = \sqrt{100} = 10$ cm. For the second chord the half-length is $\sqrt{10^2 - 8^2} = \sqrt{36} = 6$ cm, so the chord is $12$ cm. (Pythagoras twice: first for the radius, then for the chord.)

A tangent is always perpendicular to the radius at the point of contact, so the angle is $90^\circ$. (This fact lets you use Pythagoras with tangents and radii.)

Angles in the same segment, standing on the same chord $AB$, are equal: $\angle ADB = \angle ACB = 35^\circ$.

By the cosine rule in the isosceles triangle of two radii, $c^2 = 10^2 + 10^2 - 2(10)(10)\cos 120^\circ = 200 - 200(-\tfrac{1}{2}) = 300$, so $c = 10\sqrt{3}$ cm. (Equivalently the perpendicular from the centre gives $2 \times 10\sin 60^\circ$.)

Opposite angles are supplementary. Let $\angle C = x$; then $\angle A = 2x$ and $2x + x = 180$, so $x = 60$ and $\angle A = 120^\circ$. (The cyclic-quadrilateral theorem combined with solving an equation.)

The perpendicular from the centre bisects the chord, so $AM = 8$ cm. In right triangle $OMA$: $OM = \sqrt{10^2 - 8^2} = \sqrt{36} = 6$ cm.

For intersecting chords, $AE \times EB = CE \times ED$, so $4 \times 6 = 3 \times ED$, giving $ED = \tfrac{24}{3} = 8$.

Since $AB$ is a diameter, $\angle ACB = 90^\circ$ (angle in a semicircle). The angles of triangle $ABC$ sum to $180^\circ$, so $\angle BAC = 180^\circ - 90^\circ - 35^\circ = 55^\circ$.

Radii are equal, so the base angles are equal: $\angle OBA = \angle OAB = 40^\circ$. Then $\angle AOB = 180^\circ - 40^\circ - 40^\circ = 100^\circ$.

The alternate segment theorem states the tangent-chord angle equals the inscribed angle in the alternate segment, so $\angle ACB = 65^\circ$. (25° wrongly took the complement.)

The angle in a semicircle gives $\angle ACB = 90^\circ$, so $\angle BAC = 180^\circ - 90^\circ - 55^\circ = 35^\circ$ (angle sum). Then $\angle BOC = 2\,\angle BAC = 70^\circ$ (angle at the centre is twice the angle at the circumference on arc $BC$). (Three theorems chained.)

The tangent is perpendicular to the radius at $A$, so triangle $OAP$ is right-angled at $A$: $PA = \sqrt{13^2 - 5^2} = \sqrt{144} = 12$ cm (a 5-12-13 triangle).

A tangent is perpendicular to the radius, so $\angle OAP = \angle OBP = 90^\circ$. The angles of quadrilateral $OAPB$ sum to $360^\circ$: $\angle APB = 360^\circ - 90^\circ - 90^\circ - 140^\circ = 40^\circ$. (Tangent-radius perpendicularity plus the angle sum of a quadrilateral.)

The diameter is $AB = \sqrt{(4 - (-2))^2 + (9 - 1)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$. The radius is half of that, $5$. (Distance formula, then the diameter-radius relationship.)

The tangents are equal, so the base angles are equal: $\angle PAB = \angle PBA = \tfrac{180^\circ - 40^\circ}{2} = 70^\circ$.

The exterior angle of a cyclic quadrilateral equals the interior opposite angle, so it is $110^\circ$. (The $70^\circ$ is the adjacent interior angle, not the opposite one.)

By the alternate segment theorem, $\angle ACB = 50^\circ$. Since $CA = CB$, triangle $ABC$ is isosceles, so $\angle ABC = \angle BAC = \tfrac{180^\circ - 50^\circ}{2} = 65^\circ$. (Alternate segment theorem plus isosceles-triangle angles.)

The perpendicular from the centre to a chord passes through its midpoint, $(3, 3)$. The distance is $\sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2}$. (Midpoint and distance formulas with the chord-bisection property.)

Each side subtends $120^\circ$ at the centre (a third of $360^\circ$), so by the cosine rule the side is $\sqrt{6^2 + 6^2 - 2(6)(6)\cos 120^\circ} = \sqrt{108} = 6\sqrt{3}$ cm. (Equivalently, side $= R\sqrt{3}$ for a triangle inscribed in a circle of radius $R$.)

$OA = OB$ (radii), so triangle $OAB$ is isosceles and $\angle AOB = 180^\circ - 2(25^\circ) = 130^\circ$. Then $\angle ACB = \tfrac{1}{2}\angle AOB = 65^\circ$ (angle at the centre is twice the angle at the circumference). (Isosceles triangle, then the centre-circumference theorem.)

Opposite angles are supplementary: $(3x - 10) + (x + 30) = 180$, so $4x + 20 = 180$ and $x = 40$. Then $\angle A = 3(40) - 10 = 110^\circ$. (Set up the equation from the theorem, solve, then substitute back.)

The square's diagonal is the diameter, $20$ cm. For a square, diagonal $= s\sqrt{2}$, so $s = \dfrac{20}{\sqrt{2}} = 10\sqrt{2}$ cm. (The diagonal equals the diameter, then Pythagoras / the square-diagonal relationship.)