Solutions: Circles - Test 2

Q1. (b) 10 cm

The distance between two parallel tangents is the diameter of the circle. Diameter = 2 × radius = 2 × 5 cm = 10 cm.

Q2. (d) 3 cm

The sides 8 cm, 15 cm, 17 cm form a right-angled triangle since \(8^2 + 15^2 = 64 + 225 = 289 = 17^2\).
The radius of the incircle of a right-angled triangle is \(r = (a+b-c)/2\), where c is the hypotenuse.
\(r = (8+15-17)/2 = 6/2 = 3\) cm.

Q3. (b) 2a

Let the tangents be PA and PB. In \(\triangle OAP\), \(\angle OAP = 90^\circ\). The line OP bisects the angle between the tangents, so \(\angle APO = 60^\circ/2 = 30^\circ\).
\(\sin(30^\circ) = OA/OP = a/OP \Rightarrow 1/2 = a/OP \Rightarrow OP = 2a\).

Q4. (b) Both A and R are true but R is not the correct explanation of A.

Assertion (A) is a fundamental theorem. Reason (R) is also a fundamental theorem. However, (R) does not explain (A). They are independent properties of tangents.

Q5. Prove that \(XA + AR = XB + BR\).

We know that the lengths of tangents drawn from an external point to a circle are equal.
From external point A, tangents are to points P and R. Thus, \(AP = AR\). (1 Mark)
From external point B, tangents are to points Q and R. Thus, \(BQ = BR\). (1 Mark)
From external point X, tangents are XP and XQ. Thus, \(XP = XQ\).
We can write \(XP = XA + AP\) and \(XQ = XB + BQ\).
Substituting \(AP = AR\) and \(BQ = BR\), we get \(XP = XA + AR\) and \(XQ = XB + BR\).
Since \(XP = XQ\), it follows that \(XA + AR = XB + BR\). Hence proved.

Q6. Prove that \(\angle PTQ = 2\angle OPQ\).

Let \(\angle PTQ = \theta\). Since TP = TQ (tangents from T), \(\triangle PTQ\) is isosceles.
\(\angle TPQ = \angle TQP = (180^\circ - \theta)/2 = 90^\circ - \theta/2\). (1 Mark)
We know that the radius is perpendicular to the tangent at the point of contact, so \(\angle OPT = 90^\circ\).
\(\angle OPQ = \angle OPT - \angle TPQ = 90^\circ - (90^\circ - \theta/2) = \theta/2\>. So, \(\angle OPQ = \theta/2 = (\angle PTQ)/2\), which implies \(\angle PTQ = 2\angle OPQ\). (1 Mark)

Q7. Find the length of \(BC\).

Given \(AB = 10\) cm, \(AR = 7\) cm, \(CR = 5\) cm.
Tangents from an external point are equal: \(AP = AR = 7\) cm, \(CQ = CR = 5\) cm, \(BP = BQ\).
\(BP = AB - AP = 10 - 7 = 3\) cm. (1 Mark)
Therefore, \(BQ = 3\) cm.
\(BC = BQ + CQ = 3 + 5 = 8\) cm. BC = 8 cm (1 Mark)

Q8. Find the length \(TP\).

Let O be the center. OT bisects chord PQ at M, so \(PM = 4\) cm and \(\angle OMP = 90^\circ\>. In \(\triangle OMP\), \(OM = \sqrt{OP^2 - PM^2} = \sqrt{5^2 - 4^2} = \sqrt{9} = 3\) cm. (1 Mark)
In right-angled \(\triangle OPT\), \(\angle OPT = 90^\circ\>. \(\triangle OMP\) is similar to \(\triangle OPT\>. Therefore, \(TP/OP = PM/OM\). \(TP/5 = 4/3 \Rightarrow TP = 20/3\) cm. TP = 20/3 cm (2 Marks)

Q9. Prove that the parallelogram circumscribing a circle is a rhombus.

Let the parallelogram be ABCD. Let the points of contact be P, Q, R, S.
We know \(AP=AS, BP=BQ, CR=CQ, DR=DS\). (1 Mark)
Adding these gives \((AP+BP)+(CR+DR) = (AS+DS)+(BQ+CQ)\), which is \(AB+CD = AD+BC\). (1 Mark)
Since ABCD is a parallelogram, \(AB=CD\) and \(AD=BC\). Substituting gives \(2AB = 2AD \Rightarrow AB=AD\). A parallelogram with equal adjacent sides is a rhombus. Hence proved. (1 Mark)

Q10. Find the sides \(AB\) and \(AC\).

Let the circle touch AB at F and AC at E. Given radius \(r=4\). \(CD=CE=6\) cm, \(BD=BF=8\) cm. Let \(AE=AF=x\). Sides are \(a=14\), \(b=x+6\), \(c=x+8\). (1 Mark)
Semi-perimeter \(s = (14+x+6+x+8)/2 = 14+x\). Area of \(\triangle ABC\) by Heron's formula is \(\sqrt{s(s-a)(s-b)(s-c)} = \sqrt{(14+x)(x)(8)(6)} = \sqrt{48x(14+x)}\). (2 Marks)
Area is also \(r \times s = 4(14+x)\). (1 Mark)
Equating the areas: \(\sqrt{48x(14+x)} = 4(14+x)\). Squaring both sides: \(48x(14+x) = 16(14+x)^2 \Rightarrow 3x = 14+x \Rightarrow 2x=14 \Rightarrow x=7\). Sides are \(AB = x+8 = 15\) cm and \(AC = x+6 = 13\) cm. AB=15 cm, AC=13 cm (1 Mark)

Q11. Case Study:

(i) Distance of the point on the ground (P) from the center of the wheel (O). We have a right triangle with height 12 m and base 24 m. \(OP = \sqrt{12^2 + 24^2} = \sqrt{144 + 576} = \sqrt{720} = 12\sqrt{5}\) m. \(12\sqrt{5}\) m (1 Mark)

(ii) Length of the safety beam (tangent) from P to the point of contact T. Using Pythagoras in \(\triangle OTP\), \(PT = \sqrt{OP^2 - r^2} = \sqrt{720 - 10^2} = \sqrt{620} = 2\sqrt{155}\) m. \(2\sqrt{155}\) m (2 Marks)

(iii) Angle of elevation \(\theta\). \(\tan \theta = \text{opposite}/\text{adjacent} = 12/24 = 1/2\). \(\tan \theta = 1/2\) (1 Mark)