CAT Tangent to a Circle - Practice Questions & MCQ

A point inside the circle: 

Let’s draw a circle on a paper and mark a point P inside the circle. From this point try to  draw a tangent to the circle, as you see the lines through this point P intersect the circles at two points and you did not find a single tangent. 

A point on the circle: 

Now take point P on the circle and try to draw a tangent through this point. You see that you will  get  only one tangent PT through the point P. 

A point outside the circle: 

Now take the point P outside the circle and from this point try to draw the tangents to  the given circle. Here you get two tangents PT and PQ from the point P to the given circle. 

The number of tangents drawn from a point to a circle depends upon the position of the  point with respect to the circle. 

1. If a point lies inside the circle, any line passing through the point will intersect the circle at two points and  therefore, cannot be a tangent.
2. If a point  lies on a circle, only one tangent can be drawn to the circle through the point of  contact.
3. If point lies outside the circle, two tangents can be drawn to the circle from that point.

Theorem 1: The tangent at any point of a circle is perpendicular to the radius through the point of contact.

Let's prove this

Given, A circle C (O, r) and a tangent AB at a point P. 

Take any point Q, other than P, on the tangent AB. Join OQ.

Since, Q is a point on the tangent AB, other than the point of contact P, so Q will be outside the circle.  Let OQ intersect the circle at R. 

         OQ = OR + RQ  

⇒       OQ > OR  

⇒       OQ > OP     (OR = OP = radius)  

ஃ        OP < OQ 

Thus, OP is shorter than any other segment joining O to any point of AB. 

But, among all line segments, joining the point O to a point on AB, the shortest one is the perpendicular from O  on AB. 

Hence, OP ⊥ AB.

By the above, theorem,  we can also conclude that at any point on a circle there can be one and only one tangent.

The length of a tangent is the length of the segment of the tangent between the point and the given point of contact on the circle. 

Theorem 2: The lengths of tangents drawn from an external point to a circle, are  equal. 

Proof

Given. AP and AQ are two tangents, from A to a circle C (O, r) 

Join OP, OQ and OA. 

In ΔOPA and ΔOQA 

OP = OQ                   (Radii of the same circle)  

∠OPA = ∠OQA           (OP ⊥ AP and OQ ⊥ AQ)  

OA = OA                   (Common)  

ΔOPA = ΔOQA           (R.H.S. Theorem of congruence)  

AP = AQ

The above theorem can also be proved by using the Pythagoras’ theorem:

Theorem 3: A quadrilateral ABCD is drawn to circumscribe a circle, as shown in the figure. Then AB + CD = AD + BC.

Proof: We know that the lengths of tangents drawn from an exterior point to a circle are equal.

∴  AP = AS                ....(1)    [tangents from A]

    BP = BQ                ....(2)    [tangents from B]

    CR = CQ                ....(3)    [tangents from C]

    DR = DS                ....(4)    [tangents from D]

∴  AB + CD = (AP + BP) + (CR + DR) 

                   = (AS + BQ) + (CQ + DS)    [from (1), (2), (3) and (4)]

                   = (AS + DS) + (BQ + CQ)

                   = AD + BC.

Hence, AB + CD = AD + BC.

Theorem 4: Alternate Segment Theorem

The alternate segment theorem (also known as the tangent-chord theorem) states that in any circle, the angle between a chord and a tangent through one of the endpoints of the chord is equal to the angle in the alternate segment.

In the above diagram, the angles of the same color are equal to each other. 

AB is a chord in a circle with center O. A tangent is drawn to the circle at A. Chord AB makes two angles with the tangents ∠BAY and ∠BAX. Chord AB divides the circle into two segments ACB and ADB. The segments ACB and ADB are called alternate segments to angles ∠BAY and ∠BAX respectively.

Let us prove this

XY is a tangent to the given circle with center O at the point A, which lies in between X and Y. AB is a chord. C and D are points on the circle on either side of line AB.

We have to prove ∠BAY = ∠ACB and ∠BAX = ∠ADB.

Draw the diameter AOP and join PB.

∠ACB = ∠APB           (Angles in the same segment)

∠ABP = 90°               (Angle in a semi-circle)

In the triangle ABP,

∠APB + ∠BAP = 90°                   ...(1)

∠PAY = 90°               (the radius makes a right angle with the tangent at the point of tangency).

⇒ ∠BAP + ∠BAY = 90°         ...(2)

From the Eqs. (1) and (2), we get

∠APB = ∠BAY

⇒ ∠ACB = ∠BAY.           (∵∠APB = ∠ACB)

Similarly, it can be proved that

∠BAX = ∠ADB.