Tangents are drawn
to the circle x^{2} + y^{2} = 50 from a point ‘P’ lying on the
x-axis. These tangents meet the y-axis at points ‘P_{1}’ and ‘P_{2}’
. Possible coordinates of ‘P’ so that area of triangle PP_{1}P_{2}
is minimum, is / are.

(A) (10, 0)

(B) ( 10_{},
0)

(C) ( -10, 0)

(D) ( -10_{}, 0)

OP
= 5 OP D
PP (D
PP Hence (A), (C) are correct |

Two circles with
radii ‘r_{1}’ and ‘r_{2}’, r_{1} > r_{2} ≥ 2 , touch each other externally.
If ‘q’ be the angle
between the direct common tangents, then .

(A) _{}

(B) _{}

(C) q = sin^{-1}_{}

(D) none of these.

sina
= Hence (B) is correct. |

If the curves ax^{2}+4
xy+2y^{2}+x+y+5 =0 and ax^{2}+6xy+5y^{2}+ 2 x+ 3y + 8 =
0 intersect at four concyclic points then the value of a is

(A) 4

(B) -4

(C) 6

(D) –6

Any second degree curve passing through the intersection of the given curves is

ax^{2}
+ 4xy + 2y^{2} + x + y + 5 + l ( ax^{2} + 6xy + 5y^{2}
+2 x + 3y + 8 ) = 0

If
it is a circle, then coefficient of x^{2} = coefficient of y^{2}
and coefficient of xy = 0

a(1+ l) = 2 + 5l and 4 + 6 l = 0

&⇒ a = _{}and l
= _{}

&⇒
a = _{}= – 4 .

Hence (B) is correct answer.

The chords of
contact of the pair of tangents drawn from each point on the line 2x +
y = 4 to the circle x^{2} + y^{2} =1 pass through a fixed
point

(A) (2 , 4)

(B) _{}

(C) _{}

(D) ( -2, -4)

The chord of contact of tangents from (a, b) is

ax + b y = 1 . . . . . (1).

Also, (a, b )lies on 2x +y = 4 , so 2a + b = 4

&⇒ _{}

Hence,
(1) passes through _{}.

Hence (C) is correct answer.

Equation of chord AB of circle x^{2}
+ y^{2} = 2 passing through P(2 , 2) such that PB/PA = 3, is given by

(A) x = 3 y

(B) x = y

(C) y – 2 = _{}(x
– 2)

(D) none of these

Any line passing through (2, 2) will
be of the form _{}= r

When this line cuts the circle x^{2}+y^{2}=2
, (rcosq+2)^{2} +(r sinq+2)^{2}
=2

&⇒
r^{2} + 4(sinq+ cosq)r +6 = 0

_{}, now if r_{1}
= a, r_{2} = 3a,

then 4a = - 4(sinq
+ cosq), 3a^{2} = 6

&⇒ sin2q = 1

&⇒
q
= p/4 .

So required chord will be y – 2 = 1
( x –2)

&⇒ y = x.

PA.PB = PT^{2} = 2^{2} + 2^{2} – 2 =
6 . . . . (1).

_{} .
. . . (2)

From
(1) and (2), we have PA = _{}, PB =3 _{}

&⇒ AB = 2 _{}. Now diameter
of the circle is 2_{} (as radius is_{})

Hence
line passes through the centre

&⇒ y = x .

Hence (B) is the correct answer. .

Equation of a circle S(x , y) = 0, (S(2, 3) = 16) which touches the line 3x+ 4y –7 = 0 at (1, 1) is given by

(A) x^{2} +y^{2}
+x +2y –5 =0

(B) x^{2} +y^{2}
+2x +2y –6 =0

(C) x^{2} +y^{2}
+4x –6y =0

(D) none of these

Any circle which touches 3x +4y – 7 =0 at (1, 1) will be of the form

S(x, y) º
(x –1)^{2} + (y-1)^{2} +l(3x +4y-7) = 0

Since S(2, 3) = 16

&⇒
l
=1, so required circle will be

x^{2} +y^{2} +x +2y
–5 =0.

Hence (A) is the correct answer. .

If (a, 0) is a point
on a diameter of the circle x^{2}+y^{2} =4, then

x^{2} – 4x – a^{2} = 0 has

(A) exactly one real root in ( –1, 0]

(B) exactly one real root in [ 2, 5]

(C) distinct roots greater than –1

(D) distinct roots less than 5

Since (a, 0) is a point on the diameter of the circle x so
maximum value of a Let
f(x) = x f(2)
= -(a f(0)=
-a |

so graph of f(x) will be as shown

Hence (A), (B), (C), (D) are the correct answers.

If a circle S(x , y) = 0 touches at the point (2, 3) of the line x +y = 5 and S(1, 2) = 0, then radius of such circle

(A) 2 units

(B) 4 units

(C) _{}units

(D) _{}units

Desired equation of the circle is (x
–2)^{2} + (y –3)^{2} + l( x +y –5) = 0

1 +1 + l (1+ 2 – 5 ) =
0

&⇒ l =1

x^{2} – 4x + 4 + y^{2}
– 6y + 9 + x + y –5 = 0

&⇒ x^{2} + y^{2} – 3x –
5y + 8 = 0

_{}.

Hence (D) is the correct answer.

If P(2, 8) is an
interior point of a circle x^{2} + y^{2} –2x + 4y – p = 0
which neither touches nor intersects the axes, then set for p is

(A) p < -1

(B) p < -4

(C) p > 96

(D) f

For internal point p(2, 8), 4 + 64 –
4 + 32 – p < 0

&⇒ p > 96

and x intercept = 2 _{}therefore
1 + p < 0

&⇒
p < -1 and y intercept = 2_{}

&⇒
p < -4

Hence (D) is the correct answer.

If two
circles (x – 1)^{2 }+ (y – 3)^{2} = r^{2} and x^{2 }+
y^{2 }– 8x + 2y + 8 = 0 intersect in two distinct points then

(A) 2 < r < 8

(B) r < 2

(C) r = 2

(D) r > 2

Let d be the distance between the
centres of two circles of radii r_{1}

and r_{2} . .

These circle intersect
at two distinct points if ½r_{1}-r_{2} ½
< d < r_{1}+r_{2}

Here, the radii of the two circles are r and 3 and distance between the

centres is 5.

Thus, ½r-3½
< 5 < r+3

&⇒ -2 < r < 8 and r > 2

&⇒
2 < r < 8.

Hence (A) is the correct answer. .

The
common chord of x^{2}+y^{2}-4x-4y = 0 and x^{2}+y^{2}
= 16 subtends at the origin an angle equal to

(A) p/6

(B) p/4

(C) p/3

(D) p/2

The equation of the
common chord of the circles x^{2}+y^{2}-4x-4y = 0 and x^{2}+y^{2
}= 16 is x+y = 4 which meets the circle x^{2}+y^{2} = 16
at points A(4,0) and B(0,4). Obviously OA ^ OB. Hence the common chord AB makes
a right angle at the centre of the circle x^{2}+y^{2} = 16.

Hence (D) is the correct answer.

The
number of common tangents that can be drawn to the circle

x^{2}+y^{2}–4x-6y-3 = 0 and x^{2}+y^{2}+2x+2y+1=0
is

(A) 1

(B) 2

(C) 3

(D) 4

The two circles are

x^{2 }+ y^{2
}– 4x – 6y – 3 = 0 and x^{2}+y^{2}+2x+2y+1 = 0

Centre: C_{1} º
(2, 3), C_{2} º (–1, –1), radii: r_{1} = 4,
r_{2} = 1

We have, C_{1}
C_{2} = 5 = r_{1} + r_{2}, therefore there are 3 common
tangents to the given circles. Hence (C) is the correct answer.

The
tangents drawn from the origin to the circle x^{2}+y^{2}-2rx-2hy+h^{2}
= 0 are perpendicular if

(A) h = r

(B) h = -r

(C) r^{2
}+ h^{2} = 1

(D) r^{2
}= h^{2}

The combined equation of the tangents drawn from (0,0) to

x^{2 }+ y^{2}–
2rx – 2 hy + h^{2} = 0 is

(x^{2 }+ y^{2}
– 2 rx – 2 hy + h^{2})h^{2} = ( – rx – hy + h^{2})^{2}

This equation represents a pair of perpendicular straight lines

If Coeff. of x^{2}
+ coeff. of y^{2} = 0 i.e. 2h^{2} – r^{2} – h^{2}
= 0 .

&⇒
r^{2} = h^{2} or r = ± h. Hence (A), (B), and (D) are
correct answers.

The equation(s) of the tangent at the point (0, 0) to the circle, making intercepts of length 2a and 2b units on the coordinate axes, is (are)

(A) ax + by = 0

(B) ax – by = 0

(C) x = y

(D) None of these

Equation of circle passing through
origin and cutting off intercepts 2a and 2b units on the coordinate axes is x^{2}
+ y^{2} ± 2ax ± 2by = 0

and equation of tangent at (0, 0) is ax ± by = 0

Hence (A), (B) are correct answers.

The slope of the
tangent at the point (h,h) of the circle x^{2 }+ y^{2 }= a^{2}
is

(A) 0

(B) 1

(C) -1

(D) Depend on h

The equation of the tangent at (h, h)
to x^{2} + y^{2} = a^{2} is hx + hy = a^{2}.

Therefore slope of the tangent = -h/h = -1

Hence (C) is the correct answer.