Class 10 Mathematics - Chapter 10
Complete Solutions & Proofs for Chapter: Circles
Exercise 10.1
A circle can have infinitely many tangents.
Reasoning: A circle consists of an infinite number of points on its perimeter. Since a tangent can be drawn at every single point on the circle, there can be infinitely many tangents.
- A tangent to a circle intersects it in one point(s).
- A line intersecting a circle in two points is called a secant.
- A circle can have two parallel tangents at the most.
- The common point of a tangent to a circle and the circle is called point of contact.
By Theorem 10.1, the line segment joining the centre to the point of contact is perpendicular to the tangent line (\(OP \perp PQ\)).
Therefore, \(\triangle OPQ\) is a right-angled triangle at \(P\).
Applying the Pythagoras theorem in \(\triangle OPQ\):
\[ OQ^2 = OP^2 + PQ^2 \] \[ 12^2 = 5^2 + PQ^2 \implies 144 = 25 + PQ^2 \] \[ PQ^2 = 144 - 25 = 119 \implies PQ = \sqrt{119}\text{ cm} \]Steps:
- Draw a circle with centre \(O\) and a given reference line \(l\).
- Draw a line \(AB\) parallel to line \(l\) that touches the circle at exactly one point \(P\). Line \(AB\) is a tangent.
- Draw another line \(CD\) parallel to line \(l\) that intersects the circle at two distinct points \(M\) and \(N\). Line \(CD\) is a secant.
Exercise 10.2
Let \(P\) be the point of contact on the circle with centre \(O\). Since radius is perpendicular to tangent (\(OP \perp PQ\)):
\[ OQ^2 = OP^2 + PQ^2 \] \[ 25^2 = OP^2 + 24^2 \implies 625 = OP^2 + 576 \] \[ OP^2 = 625 - 576 = 49 \implies OP = \sqrt{49} = 7\text{ cm} \]Since \(OPTQ\) forms a quadrilateral and \(OP \perp TP\), \(OQ \perp TQ\) (\(\angle OPT = 90^\circ\), \(\angle OQT = 90^\circ\)):
The sum of interior angles of quadrilateral \(OPTQ\) is \(360^\circ\):
\[ \angle OPT + \angle POQ + \angle OQT + \angle PTQ = 360^\circ \] \[ 90^\circ + 110^\circ + 90^\circ + \angle PTQ = 360^\circ \] \[ 290^\circ + \angle PTQ = 360^\circ \implies \angle PTQ = 360^\circ - 290^\circ = 70^\circ \]In quadrilateral \(AOBP\), \(\angle OAP = 90^\circ\) and \(\angle OBP = 90^\circ\).
\[ \angle AOB = 180^\circ - \angle APB = 180^\circ - 80^\circ = 100^\circ \]Since \(\triangle POA \cong \triangle POB\), line \(OP\) bisects \(\angle AOB\):
\[ \angle POA = \frac{1}{2} \angle AOB = \frac{100^\circ}{2} = 50^\circ \]Let \(AB\) be a diameter of a circle with centre \(O\). Let \(l_1\) and \(l_2\) be the tangents drawn at points \(A\) and \(B\) respectively.
Since radius is perpendicular to the tangent at the point of contact:
\[ OA \perp l_1 \implies \angle OAB = 90^\circ \] \[ OB \perp l_2 \implies \angle OBA = 90^\circ \]Here, \(\angle OAB\) and \(\angle OBA\) form a pair of alternate interior angles for lines \(l_1\) and \(l_2\) with transversal \(AB\).
Since alternate interior angles are equal (\(90^\circ = 90^\circ\)), the lines \(l_1\) and \(l_2\) are parallel. Hence proved.
Let a tangent \(AB\) touch the circle with centre \(O\) at point \(P\).
Suppose the perpendicular to \(AB\) at point \(P\) does not pass through centre \(O\). Let it pass through another point \(O'\).
Then, by definition of perpendicularity, \(\angle O'PA = 90^\circ\) --- (1)
However, we know that the radius through the point of contact is perpendicular to the tangent (Theorem 10.1). Therefore, \(\angle OPA = 90^\circ\) --- (2)
From (1) and (2), we get \(\angle O'PA = \angle OPA\).
This is possible only when the line segment \(O'P\) coincides with \(OP\). Hence, the perpendicular at the point of contact must pass through the centre \(O\). Hence proved.
Let \(O\) be the centre and \(P\) be the point of contact on the circle. Given \(OA = 5\text{ cm}\) and \(AP = 4\text{ cm}\).
Since \(OP \perp AP\), in right \(\triangle OPA\):
\[ OA^2 = OP^2 + AP^2 \] \[ 5^2 = r^2 + 4^2 \implies 25 = r^2 + 16 \] \[ r^2 = 25 - 16 = 9 \implies r = 3\text{ cm} \]The radius of the circle is \(3\text{ cm}\).
Let \(O\) be the common centre. Let \(AB\) be the chord of the outer circle (radius \(R = 5\text{ cm}\)) touching the inner circle (radius \(r = 3\text{ cm}\)) at point \(P\).
\(OP \perp AB\) because \(AB\) is a tangent to the inner circle at \(P\).
In right \(\triangle OPA\):
\[ OA^2 = OP^2 + AP^2 \implies 5^2 = 3^2 + AP^2 \] \[ 25 = 9 + AP^2 \implies AP^2 = 16 \implies AP = 4\text{ cm} \]Since the perpendicular from the centre to a chord bisects the chord, \(AB = 2 \times AP = 2 \times 4 = 8\text{ cm}\).
The length of the chord is \(8\text{ cm}\).
Let the circle touch the sides \(AB, BC, CD,\) and \(DA\) at points \(P, Q, R,\) and \(S\) respectively.
By Theorem 10.2, lengths of tangents drawn from an external point are equal:
- \(AP = AS\) --- (1)
- \(BP = BQ\) --- (2)
- \(CR = CQ\) --- (3)
- \(DR = DS\) --- (4)
Adding equations (1), (2), (3), and (4):
\[ (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ) \] \[ AB + CD = AD + BC \]Hence proved.
Join \(OC\). Let \(P\) and \(Q\) be the points of contact of parallel lines \(XY\) and \(X'Y'\) respectively.
In \(\triangle OPA\) and \(\triangle OCA\):
- \(AP = AC\) (Tangents from external point \(A\))
- \(OP = OC\) (Radii of the same circle)
- \(OA = OA\) (Common)
By SSS congruency, \(\triangle OPA \cong \triangle OCA \implies \angle POA = \angle COA = \theta_1\).
Similarly, \(\triangle OQB \cong \triangle OCB \implies \angle QOB = \angle COB = \theta_2\).
Since \(POQ\) is a straight line diameter, the total angle is \(180^\circ\):
\[ \angle POA + \angle COA + \angle COB + \angle QOB = 180^\circ \] \[ 2\theta_1 + 2\theta_2 = 180^\circ \implies \theta_1 + \theta_2 = 90^\circ \]Since \(\angle AOB = \angle COA + \angle COB = \theta_1 + \theta_2 = 90^\circ\).
Hence proved.
Let \(PA\) and \(PB\) be tangents from point \(P\) to a circle with centre \(O\), touching at \(A\) and \(B\).
Since radius is perpendicular to tangent: \(\angle OAP = 90^\circ\) and \(\angle OBP = 90^\circ\).
In quadrilateral \(OAPB\), the sum of all angles is \(360^\circ\):
\[ \angle APB + \angle OAP + \angle AOB + \angle OBP = 360^\circ \] \[ \angle APB + 90^\circ + \angle AOB + 90^\circ = 360^\circ \] \[ \angle APB + \angle AOB = 360^\circ - 180^\circ = 180^\circ \]Thus, \(\angle APB\) and \(\angle AOB\) are supplementary. Hence proved.
Let \(ABCD\) be a parallelogram circumscribing a circle.
From Q8, we know that for any quadrilateral circumscribing a circle:
\[ AB + CD = AD + BC \quad \text{--- (1)} \]Since \(ABCD\) is a parallelogram, opposite sides are equal: \(AB = CD\) and \(AD = BC\).
Substituting these into (1):
\[ AB + AB = AD + AD \implies 2 AB = 2 AD \implies AB = AD \]Since adjacent sides are equal (\(AB = AD\)) and opposite sides are equal, all four sides are equal (\(AB = BC = CD = DA\)).
Therefore, \(ABCD\) is a rhombus. Hence proved.
Let the circle touch sides \(BC, CA,\) and \(AB\) at \(D, E,\) and \(F\) respectively.
Given \(CD = 6\text{ cm} \implies CE = 6\text{ cm}\) and \(BD = 8\text{ cm} \implies BF = 8\text{ cm}\).
Let \(AF = AE = x\text{ cm}\).
Sides of \(\triangle ABC\):
- \(a = BC = 6 + 8 = 14\text{ cm}\)
- \(b = AC = x + 6\text{ cm}\)
- \(c = AB = x + 8\text{ cm}\)
Semi-perimeter \(s = \frac{a+b+c}{2} = \frac{14 + (x+6) + (x+8)}{2} = \frac{2x + 28}{2} = x + 14\text{ cm}\).
Area using Heron's Formula:
\[ \text{Area}(\triangle ABC) = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{(x+14)(x)(8)(6)} = \sqrt{48x(x+14)} \]Area using Inradius (\(r = 4\text{ cm}\)):
\[ \text{Area}(\triangle ABC) = r \times s = 4(x + 14) \]Equating both area formulas and squaring both sides:
\[ 48x(x+14) = 16(x+14)^2 \] \[ 3x = x + 14 \implies 2x = 14 \implies x = 7\text{ cm} \]Therefore, the side lengths are:
- \(AB = x + 8 = 7 + 8 = 15\text{ cm}\)
- \(AC = x + 6 = 7 + 6 = 13\text{ cm}\)
Let quadrilateral \(ABCD\) circumscribe a circle with centre \(O\), touching at \(P, Q, R, S\).
Join \(OP, OQ, OR, OS\) and vertices \(OA, OB, OC, OD\).
This divides the total angle around the centre into 8 smaller angles \(\angle 1, \angle 2, \dots, \angle 8\).
Using congruent triangles (e.g., \(\triangle OAP \cong \triangle OAS\)):
\[ \angle 1 = \angle 8, \quad \angle 2 = \angle 3, \quad \angle 4 = \angle 5, \quad \angle 6 = \angle 7 \]Since sum of angles around a point is \(360^\circ\):
\[ (\angle 1 + \angle 2 + \angle 3 + \angle 4 + \angle 5 + \angle 6 + \angle 7 + \angle 8) = 360^\circ \] \[ 2(\angle 2 + \angle 3 + \angle 6 + \angle 7) = 360^\circ \implies (\angle 2 + \angle 3) + (\angle 6 + \angle 7) = 180^\circ \] \[ \angle AOB + \angle COD = 180^\circ \]Similarly, \(\angle AOD + \angle BOC = 180^\circ\). Hence proved.
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