Circles: Tangents, Secants and Intersecting Chords — chapter notes

• A tangent meets a circle at exactly one point, the point of contact; a secant meets it at two points and a chord is the part of a secant inside the circle.
• Theorem: the tangent at a point is perpendicular to the radius drawn to the point of contact. Proof: every other point $X$ of the tangent is outside the circle, so $OX > OT$; $OT$ is the shortest segment from $O$ to the line, so it is perpendicular to it.
• Use it to get a right-angled triangle $OTP$: $PT = \sqrt{OP^2 - r^2}$.

Example: $OT = 20$ cm and $OP = 29$ cm give $PT = \sqrt{29^2 - 20^2} = 21$ cm; with $\angle POT = 62^\circ$, $\angle OPT = 180^\circ - 90^\circ - 62^\circ = 28^\circ$.

• The two tangents drawn to a circle from an external point are equal: $PA = PB$. Proof: the right-angled triangles $OAP$ and $OBP$ have $OA = OB$, the common side $OP$ and right angles at $A$ and $B$, so they are congruent (RHS).
• $OP$ bisects $\angle APB$ and $\angle AOB$, and $OP \perp AB$; in the quadrilateral $OAPB$: $\angle AOB = 180^\circ - \angle APB$.
• No tangent can be drawn from a point inside a circle.

Example: $\angle APB = 76^\circ$ gives $\angle AOB = 360^\circ - 90^\circ - 90^\circ - 76^\circ = 104^\circ$, $\angle OAB = \dfrac{180^\circ - 104^\circ}{2} = 38^\circ$ and $\angle PAB = 90^\circ - 38^\circ = 52^\circ$. For $PA = (3x - 4)$ cm and $PB = (x + 8)$ cm: $3x - 4 = x + 8$, $x = 6$.

• Two circles touch when they meet at exactly one point; that point and the two centres lie on one straight line (the line of centres), and the common tangent at the point is perpendicular to it.
• Externally: the centres are on opposite sides of the point of contact and $d = r_1 + r_2$.
• Internally: the centres are on the same side of the point of contact and $d = r_1 - r_2$ (the smaller circle is inside).

Example: radii $3$ and $2$ cm touching externally are $3 + 2 = 5$ cm apart; radii $5$ and $2$ cm touching internally are $5 - 2 = 3$ cm apart. If $x + (x + 5) = 25$ then $x = 10$ and the radii are $10$ cm and $15$ cm.

• A common tangent touches both circles. A direct one leaves both circles on the same side; a transverse one crosses the line of centres between the circles.
• Draw the radii to the points of contact and a line parallel to the tangent: the right-angled triangle has hypotenuse $d$ and a leg $r_1 - r_2$ (direct) or $r_1 + r_2$ (transverse).
• Direct: $\sqrt{d^2 - (r_1 - r_2)^2}$. Transverse: $\sqrt{d^2 - (r_1 + r_2)^2}$, which needs $d > r_1 + r_2$.

Example: $d = 25$, $r_1 = 11$, $r_2 = 4$: direct $\sqrt{25^2 - 7^2} = 24$, transverse $\sqrt{25^2 - 15^2} = 20$. For $r_1 = 18$, $r_2 = 3$ and $d = 25$ the direct tangent is $\sqrt{25^2 - 15^2} = 20$.

• Theorem: if two chords $AB$ and $CD$ cut at $P$ inside a circle, $PA \times PB = PC \times PD$. Proof: $\triangle PAC \sim \triangle PDB$ (equal angles in the same segment), so $\dfrac{PA}{PD} = \dfrac{PC}{PB}$.
• If the chords are produced and meet at $P$ outside the circle, the same rule holds with $PB$ and $PD$ measured from $P$ to the far ends.
• Problems with algebra give a quadratic: reject the negative root, because a length is positive.

Example: $PA = 15$, $PB = 4$, $PC = 5$: $PD = \dfrac{15 \times 4}{5} = 12$ cm. For $PA = 2x$, $PB = x + 1$, $PC = 2$, $PD = 12$: $x^2 + x - 12 = 0$, $x = 3$. For secants, $PA = 5$, $AB = 7$, $PC = 6$ give $PD = \dfrac{5 \times 12}{6} = 10$ and $CD = 10 - 6 = 4$ cm.

• Theorem: if a tangent $PT$ and a secant $PAB$ ($A$ nearer to $P$) are drawn from $P$, then $PT^2 = PA \times PB$. Proof: $\triangle PTA \sim \triangle PBT$ (common angle at $P$ and $\angle PTA = \angle PBT$), so $\dfrac{PT}{PB} = \dfrac{PA}{PT}$.
• The same square gives $PA \times PB = PC \times PD$ for two secants from the same point.
• If the secant passes through the centre, $AB$ is a diameter.

Example: $PT = 20$, $PA = 16$: $PB = \dfrac{400}{16} = 25$ and $AB = 25 - 16 = 9$ cm. For $PT = 9$, $PA = x$, $AB = 24$: $x(x + 24) = 81$, $x = 3$.

• Theorem: the angle between a tangent and a chord through the point of contact equals the angle in the alternate segment (on the other side of the chord): $\angle PTA = \angle TBA$.
• Proof: draw the diameter $TD$: $\angle PTD = 90^\circ$ and $\angle TAD = 90^\circ$, so $\angle TDA = 90^\circ - \angle ATD = \angle PTA$, and $\angle TDA = \angle TBA$ (same segment).
• It gives all three angles of a triangle in a circle from the tangent at one vertex: with $\angle PAB = 68^\circ$ and $\angle QAC = 47^\circ$, $\angle BAC = 180^\circ - 68^\circ - 47^\circ = 65^\circ$.

Example: a tangent-chord angle of $48^\circ$ gives an angle of $48^\circ$ in the alternate segment and $2 \times 48^\circ = 96^\circ$ at the centre.