Mathematics_Today_-_October_2016

Area of the triangle =^1
2

absinC

=+{}⋅ = +

⎛

⎝⎜

⎞

⎠⎟

1

2

312

3

33

2

()sinπ cm^22 cm

20. (b) : Let AB = 3, BC = 5, CD = 2, DA = x and
    ∠ABC = 60° of the cyclic quadrilateral ABCD
    ∴ ∠CDA = 180° – 60° = 120°

From ΔABC, we have cos∠ABC=AB +AB BCBC⋅ −AC

22 2

2

⇒ 1
2

925

235

2

= + −

⋅⋅

AC ⇒ AC (^2) = 19 ... (1)
From ΔADC, we have cos∠ = + −
⋅
CDA CD AD AC
CD AD
222
2
⇒ − = + −
⋅⋅
1
2
419
22
x^2
x
()Using (1)
⇒ x^2 + 2x – 15 = 0 ⇒ (x + 5)(x – 3) = 0
⇒ x = 3 (∵ x ≠ –5)

21. (d) : For the circle, x^2 + y^2 + 6x + 6y = 0,
    the centre is at (–3, –3) and radius = 32
    Also the centre and radius of the circle
    x^2 + y^2 – 12x – 12y = 0 is (6, 6) and 62 respectively.
    Distance between the centres of the circles

=+= =81 81 9 2 sum of the radii
∴ the two circles touch each other externally.

22. (d) : The equation of the circle whose diameter's

end points are (0, 0) and a
a

3

3

⎛ , 1

⎝⎜

⎞

⎠⎟ is

()( )()xxay y

a

−−+ −−⎛

⎝⎜

⎞

⎠⎟

003 13 = 0

.... (1)

Since the equation (1) is satisfied by^1
a

⎛⎝⎜ ,a⎞⎠⎟, therefore

the circle is passing through^1
a

⎛⎝⎜ ,a⎞⎠⎟.

23. (d) : Hint: Two circles touch each other if the
    distance between their centres is equal to the sum or
    difference of the their radii.


24. (d) : Equation of the circle passing through the
    points (0, 0), (1, 0), (0, –1) is
    ⇒ x^2 + y^2 – x + y = 0 ... (1)
    (0, 0), (1, 0), (0, –1) and (λ, 3λ) are concyclic, therefore
    from (1), we get
    λ^2 + 9λ^2 – λ + 3λ = 0 ⇒ 10 λ^2 + 2λ = 0

⇒ 2 λ(5λ + 1) = 0 ⇒λ=−^1
5

(∵ λ ≠ 0)

25. (b) : The centre and radius of the circle x^2 + y^2 = 16
    are respectively (0, 0) and 4.
    Now the distance of the point (9, –12) from (0, 0) is
    81 144+= =^22515.
    Hence, the least distance of the circle x^2 + y^2 = 16 from
    the point (9, –12) is 15 – 4 = 11 units


26. (b) : Here, the given equation of the circle is
    x^2 + y^2 – 8x – 4y + 15 = 0
    ∴ Centre is (4, 2). Given that one extremity of the
    diameter is (2, 1). Let the other extremity be (x, y).
    Therefore the other extremity of the diameter is (6, 3)
    [Since the centre is the mid-point of diameter]


27. (c) : The centre of the circle x^2 + y^2 = 25 is (0, 0).
    Now slope of QO × slope of RO

= −

−

× −

−−

(^40) =×− =−
30
30
40
4
3
3
4
1
Therefore angle at the centre (),∠QOR=π
2
∴ Angle at the circumference is
∠QPR=×∠QOR=
1
24
π

28. (a) : Let the equation of the circle be
    (x – α)^2 + (y)^2 = α^2
    (∵ the circle touches y-axis at origin. ∴ centre is on
    x-axis)
    ⇒ x^2 – 2αx + α^2 + y^2 = α^2
    ⇒ x^2 – 2αx + y^2 = 0 ... (1)
    Since (1) is passing through (h, k), therefore

α=

hk+

h

22

2

Now (1) becomes

xy hk

h

(^22) xhxyhkx
(^22) 22 22
2
2
+= +
⎛
⎝
⎜
⎞
⎠
⎟ ⇒ ()()+=+

29. (d) : The given equation is
    ax^2 + (2a – 3)y^2 – 6x + ay – 3 = 0
    This equation will represent a circle if the coefficient of
    x^2 is equal to the coefficient of y^2.
    ∴ a = 2a – 3 ⇒ a = 3.
    Now, the equation of the circle is
    3 x^2 + 3y^2 – 6x + 3y – 3 = 0
    ⇒ x^2 + y^2 – 2x + y – 1 = 0

∴ Radius = 1++=^1

4

1 3

2

.