Ch 10: Circles (Mega Proofs)
This page contains the advanced 4-5 marker proofs from Exercise 10.2 (Q8 to Q13). Every single theorem relies heavily on Theorem 10.2.
Exercise 10.2 Solutions (Advanced)
💡 Core Logic for all proofs: The lengths of tangents drawn from an external point to a circle are EQUAL.
Q8. A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that AB + CD = AD + BC.
Proof:
Let the points of contact of the circle with sides AB, BC, CD, and DA be P, Q, R, and S respectively.
Lengths of tangents drawn from an external point are equal. Therefore:
From A: $\quad AP = AS$ --- (1)
From B: $\quad BP = BQ$ --- (2)
From C: $\quad CR = CQ$ --- (3)
From D: $\quad DR = DS$ --- (4)
Adding equations (1), (2), (3), and (4):
$(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)$
Looking at the figure, combining the segments gives the full sides:
$AB + CD = AD + BC$
Hence Proved.
Q9. In Fig. 10.13, XY and X'Y' are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X'Y' at B. Prove that $\angle AOB = 90^\circ$.
Proof:
Join OC. Let P and Q be the points of contact of parallel tangents XY and X'Y' respectively. PQ is the diameter.
In $\Delta OPA$ and $\Delta OCA$:
$OP = OC$ (Radii of same circle)
$AP = AC$ (Tangents from external point A)
$OA = OA$ (Common side)
By SSS congruence, $\Delta OPA \cong \Delta OCA$.
Therefore, $\angle POA = \angle COA$ --- (1)
Similarly, $\Delta OQB \cong \Delta OCB$.
Therefore, $\angle QOB = \angle COB$ --- (2)
Since POQ is a straight line (diameter), the sum of angles is $180^\circ$:
$\angle POA + \angle COA + \angle COB + \angle QOB = 180^\circ$
Using (1) and (2):
$2\angle COA + 2\angle COB = 180^\circ$
$2(\angle COA + \angle COB) = 180^\circ$
$\angle COA + \angle COB = 90^\circ$
Since $\angle AOB = \angle COA + \angle COB$, we get $\angle AOB = 90^\circ$. (Hence Proved)
Q10. Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.
Proof:
Let PA and PB be two tangents drawn from point P to a circle with centre O. We need to prove $\angle APB + \angle AOB = 180^\circ$.
Radius is perpendicular to the tangent at the point of contact. So, $\angle OAP = 90^\circ$ and $\angle OBP = 90^\circ$.
In quadrilateral $OAPB$, the sum of interior angles is $360^\circ$:
$\angle OAP + \angle OBP + \angle APB + \angle AOB = 360^\circ$
$90^\circ + 90^\circ + \angle APB + \angle AOB = 360^\circ$
$180^\circ + \angle APB + \angle AOB = 360^\circ$
$\angle APB + \angle AOB = 360^\circ - 180^\circ = 180^\circ$.
Since their sum is $180^\circ$, they are supplementary. (Hence Proved)
Q11. Prove that the parallelogram circumscribing a circle is a rhombus.
Proof:
Let ABCD be a parallelogram circumscribing a circle. We proved in Q8 that for any circumscribed quadrilateral:
$AB + CD = AD + BC$ --- (1)
Since ABCD is a parallelogram, opposite sides are equal:
$AB = CD$ and $AD = BC$ --- (2)
Substitute (2) into (1):
$AB + AB = AD + AD$
$2AB = 2AD \implies AB = AD$
Adjacent sides of the parallelogram are equal. Since $AB=CD$, $AD=BC$, and $AB=AD$, all four sides are equal ($AB=BC=CD=DA$).
A parallelogram with all sides equal is a rhombus. (Hence Proved)
Q12. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively. Find the sides AB and AC.
Solution:
Let the circle touch AB at F and AC at E. Tangents from external points are equal:
$CD = CE = 6$ cm
$BD = BF = 8$ cm
Let $AF = AE = x$ cm.
Sides of $\Delta ABC$:
$a = BC = 6 + 8 = 14$
$b = AC = 6 + x$
$c = AB = 8 + x$
Semi-perimeter, $s = \frac{a+b+c}{2} = \frac{14 + 6+x + 8+x}{2} = \frac{28+2x}{2} = 14 + x$.
Method 1: Area by Heron's Formula
Area = $\sqrt{s(s-a)(s-b)(s-c)}$
= $\sqrt{(14+x)(x)(8)(6)}$
= $\sqrt{48x(14+x)}$
Method 2: Area by Sum of 3 Triangles (OAB, OBC, OCA)
Area = $\frac{1}{2} \times r \times (\text{Perimeter})$
Area = $\frac{1}{2} \times 4 \times (28+2x) = 2(28+2x) = 4(14+x)$
Equating both areas and squaring both sides:
$48x(14+x) = [4(14+x)]^2$
$48x(14+x) = 16(14+x)^2$
Divide both sides by $16(14+x)$ (since $14+x \neq 0$):
$3x = 14 + x \implies 2x = 14 \implies x = 7$.
Sides are:
$AB = 8 + x = 8 + 7 = \mathbf{15 \text{ cm}}$
$AC = 6 + x = 6 + 7 = \mathbf{13 \text{ cm}}$.
Q13. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Proof:
Let ABCD be a quadrilateral circumscribing a circle with centre O. We need to prove $\angle AOB + \angle COD = 180^\circ$ and $\angle BOC + \angle AOD = 180^\circ$.
Join the centre O to all points of contact P, Q, R, S and the vertices A, B, C, D. This creates 8 small triangles.
As proved in Q9, triangles subtending from the same external point are congruent. Therefore, their central angles are equal:
$\angle 1 = \angle 2$ (From point A)
$\angle 3 = \angle 4$ (From point B)
$\angle 5 = \angle 6$ (From point C)
$\angle 7 = \angle 8$ (From point D)
The sum of all angles around centre O is $360^\circ$:
$2(\angle 2) + 2(\angle 3) + 2(\angle 6) + 2(\angle 7) = 360^\circ$
$2(\angle 2 + \angle 3 + \angle 6 + \angle 7) = 360^\circ$
$(\angle 2 + \angle 3) + (\angle 6 + \angle 7) = 180^\circ$
Since $\angle AOB = \angle 2 + \angle 3$ and $\angle COD = \angle 6 + \angle 7$, we get:
$\angle AOB + \angle COD = 180^\circ$. (Hence Proved)