Maths

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.

Properties ofTangent

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 joiningO 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.

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. 

Theorem:- If P is any point on a chord AB with centre O and radius r, then AP* PB = R2 – OP2 or AP * BP = R2 – OP2 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) = R2 -OP2 (QED)

If P lies outside :- PA * PB = PC * PD (secant property)

=> (OP-OC) (OP+ OD) = OP2 - R2

M P N

S

R <PSR = 90ºQ

(Angle in Semicircle)

<NPR = 90º (Radius to a Tangent)

<NPS + <RPS = 90º

<PRS + <RPS = 90º

<PRS + <RPS = <NPS + <RPS

<RPS is common

<PRS = <NPS

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

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

Thank You