3. Secant and Tangent A line which intersects the circle at two distinct points is called a secant of the circle. When the line intersects the circle at only one point then it is said to be a tangent .
5. Some properties of Tangent Property 1 :- A tangent to a circle is perpendicular to the circle through the point of contact. PROOF-We know that among all the line segments joining Point O on AB,the shortest one is perpendicular to AB. So we have to prove OP is shorter than any other line segment joining O to AB. OP=OR (radius of the circle) OQ= OR+ RQ. So OQ > OR. Thus OQ > OP. So OP < OQ. Hence OP is shorter than any line segment joining O to any point of AB. Hence OP is perp. To AB. Property 2 :- A line drawn through the end point of a radius and perpendicular to it is a tangent to the circle.
6. Secant property If two chords intersect inside a circle, then the product of the segments of one chord equals the product of the segments of the other chord. Chords QS and RT intersect at P . By drawing QT and RS, it can be proven that Δ QPT ∼ Δ RPS. Because the ratios of corresponding sides of similar triangles are equal, a / c = d / b . The Cross Products Property produces ( a ) ( b ) = ( c ) ( d ). This is stated as a theorem.
7. Theorem:- If P is any point on a chord AB with centre O and radius r, then AP* PB = R 2 – OP 2 or AP * BP = R 2 – OP 2 according as P is within or outside the circle. Proof :- Let CD be diameter through P. Then by Secant Property, AP * BP= CP * PD = (CO-OP) * (DP+OP) We know that CO = DO. => (CO-OP) * (DP+OP) = R 2 -OP 2 (QED) If P lies outside :- PA * PB = PC * PD (secant property) => (OP-OC) (OP+ OD) = OP 2 - R 2 Corollary
8. M P N S R <PSR = 90º Q (Angle in Semicircle) <NPR = 90º (Radius to a Tangent) <NPS + <RPS = 90º <PRS + <RPS = 90º <PRS + <RPS = <NPS + <RPS <RPS is common <PRS = <NPS Alternate Segment Theorem
10. Ptolemy's Theorem If a quadrilateral is inscribed in a circle then the sum of the products of its two pairs of opposite sides is equal to the product of its diagonals. PROOF:- Given cyclic quadrilateral ABCD as shown. From point C, make line CE such that ∠ECD ≌ ∠BCA. One can easily see that ΔABC ∽ ΔDEC which implies AB : DE = AC : DC … (1) Also because ΔCAD ∽ ΔCBE, which implies BC : AC = BE : AD … (2) By (1) : AB * DC = AC * DE By (2) : BC * AD = AC * BE Adding these two equations gives AB * DC + BC * AD = AC * DE + AC * BE = AC * (DE + BE) = AC * DB
12. Theorem:- If AD is altitude and R is circumradius of ∆ABC then AB*AC=2R*AD PROOF:- Let AE be the diameter . We have /ADC =/ABE=90 and /ACD=/AEB (angles in the same segment) Therefore, ∆ADC ║ ∆ ABE . So, AB/AD = AE/AC This gives AB*AC = AE*AD= 2R* AD
