Ch 10: Circles (Part 1)
Master the properties of Tangents, Theorem 10.1 (Perpendicularity), and Theorem 10.2 (Equal Tangents).
Exercise 10.1 Solutions
Q1. How many tangents can a circle have?
A circle can have infinitely many tangents, because a circle consists of infinite points and at every point, a unique tangent can be drawn.
Q2. Fill in the blanks:
- (i) A tangent to a circle intersects it in one point(s).
- (ii) A line intersecting a circle in two points is called a secant.
- (iii) A circle can have two parallel tangents at the most.
- (iv) The common point of a tangent to a circle and the circle is called the point of contact.
Q3. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length PQ is:
Solution:
By Theorem 10.1, the tangent is perpendicular to the radius at the point of contact. Therefore, $\angle OPQ = 90^\circ$.
In right $\Delta OPQ$, apply Pythagoras Theorem:
$OQ^2 = OP^2 + PQ^2$
$(12)^2 = (5)^2 + PQ^2$
$144 = 25 + PQ^2$
$PQ^2 = 144 - 25 = 119$
Therefore, $PQ = \sqrt{119}$ cm. Option (D) is correct.
Q4. Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.
Solution:
Draw a circle with center O. Draw any line $l$ outside the circle.
Now draw a line $m$ parallel to $l$ which just touches the circle at one point (This is the tangent).
Draw another line $n$ parallel to $l$ which cuts the circle at two distinct points (This is the secant).
Exercise 10.2 (Q1 to Q7)
💡 Trick: Most questions here use Theorem 10.1 (Radius $\perp$ Tangent) followed by Pythagoras Theorem.
Q1. From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is:
Solution:
Let O be the center and P be the point of contact. OP is the radius.
$PQ = 24$ cm, $OQ = 25$ cm. $\angle OPQ = 90^\circ$ (Radius $\perp$ Tangent).
By Pythagoras in $\Delta OPQ$:
$OQ^2 = OP^2 + PQ^2$
$25^2 = OP^2 + 24^2$
$625 = OP^2 + 576$
$OP^2 = 49 \implies \text{Radius } OP = \mathbf{7 \text{ cm}}$.
Option (A) is correct.
Q2. In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that $\angle POQ = 110^\circ$, then $\angle PTQ$ is equal to:
Solution:
Radius is perpendicular to the tangent at the point of contact. So, $\angle OPT = 90^\circ$ and $\angle OQT = 90^\circ$.
In quadrilateral $OPTQ$, the sum of interior angles is $360^\circ$:
$\angle OPT + \angle OQT + \angle POQ + \angle PTQ = 360^\circ$
$90^\circ + 90^\circ + 110^\circ + \angle PTQ = 360^\circ$
$290^\circ + \angle PTQ = 360^\circ$
$\angle PTQ = 360^\circ - 290^\circ = \mathbf{70^\circ}$.
Option (B) is correct.
Q3. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of $80^\circ$, then $\angle POA$ is equal to:
Solution:
Given $\angle APB = 80^\circ$. Line $OP$ bisects the angle between the tangents, so $\angle APO = 40^\circ$.
In right $\Delta OAP$, $\angle OAP = 90^\circ$ (Radius $\perp$ Tangent).
Sum of angles in $\Delta OAP$ = $180^\circ$:
$\angle POA + \angle APO + \angle OAP = 180^\circ$
$\angle POA + 40^\circ + 90^\circ = 180^\circ$
$\angle POA + 130^\circ = 180^\circ \implies \mathbf{\angle POA = 50^\circ}$.
Option (A) is correct.
Q4. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Proof:
Let AB be the diameter of the circle with centre O. Let lines $l$ and $m$ be tangents drawn at ends A and B respectively.
Since radius is perpendicular to tangent at point of contact:
$OA \perp l \implies \angle 1 = 90^\circ$
$OB \perp m \implies \angle 2 = 90^\circ$
Here, $\angle 1$ and $\angle 2$ form a pair of alternate interior angles. Since they are equal ($90^\circ = 90^\circ$), the lines $l$ and $m$ must be parallel.
Hence Proved.
Q5. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Proof (By Contradiction):
Let AB be a tangent to a circle at point P, and O be the centre. Let's assume the perpendicular at P to AB does NOT pass through O, but passes through another point O'.
According to our assumption: $\angle O'PB = 90^\circ$ --- (1)
But by Theorem 10.1, the radius through the point of contact is perpendicular to the tangent: $\angle OPB = 90^\circ$ --- (2)
Comparing (1) and (2), we get $\angle O'PB = \angle OPB$. This is only possible if line O'P coincides with line OP.
Therefore, the perpendicular at the point of contact must pass through the centre. (Hence Proved)
Q6. The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.
Solution:
Let O be the centre and P be the point of contact. $OA = 5$ cm, $PA = 4$ cm.
$\angle OPA = 90^\circ$ (Radius $\perp$ Tangent).
By Pythagoras Theorem in $\Delta OPA$:
$OA^2 = OP^2 + PA^2$
$5^2 = OP^2 + 4^2$
$25 = OP^2 + 16 \implies OP^2 = 9 \implies \mathbf{OP = 3 \text{ cm}}$.
The radius of the circle is 3 cm.
Q7. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Solution:
Let O be the common centre. Let AB be the chord of the larger circle which touches the smaller circle at point P.
Radius of larger circle $OA = 5$ cm. Radius of smaller circle $OP = 3$ cm.
Since AB is tangent to the smaller circle at P, $OP \perp AB \implies \angle OPA = 90^\circ$.
In right $\Delta OPA$, by Pythagoras Theorem:
$OA^2 = OP^2 + AP^2$
$5^2 = 3^2 + AP^2 \implies 25 = 9 + AP^2 \implies AP^2 = 16 \implies \mathbf{AP = 4 \text{ cm}}$.
A perpendicular drawn from the centre to a chord bisects the chord. So, $AB = 2 \times AP = 2 \times 4 = \mathbf{8 \text{ cm}}$.
The length of the chord is 8 cm.