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PASSAGE1
ABCD is a square of side length 2 units. C_{1 }is the circle touching all the sides of the square ABCD and C_{2} is the circumcircle of square ABCD. L is a fixed line in the same plane and R is a fixed point.
Q. 1. If P is any point of C_{1} and Q is another point on C_{2}, then
is equal to (2006  5M, –2)
(a) 0.75
(b) 1.25
(c)1
(d) 0.5
Ans. Sol. (a) Without loss of generality we can assume the square ABCD with its vertices A (1, 1), B (–1, 1), C (–1, –1), D (1, –1)
P to be the point (0, 1) and Q as (,0 ).
Then,
PASSAGE1
ABCD is a square of side length 2 units. C_{1 }is the circle touching all the sides of the square ABCD and C_{2} is the circumcircle of square ABCD. L is a fixed line in the same plane and R is a fixed point.
Q. 2. If a circle is such that it touches the line L and the circle C_{1 }externally, such that both the circles are on the same side of the line, then the locus of centre of the circle is (2006  5M, –2)
(a) ellipse
(b) hyper bola
(c) parabola
(d) pair of straight line
Ans. Sol. 2. (b) Let C' be the said circle touching C_{1} and L, so that C_{1} and C' are on the same side of L. Let us draw a line T parallel to L at a distance equal to the radius of circle C_{1}, on opposite side of L.
Then the centre of C' is equidistant from the centre of C_{1} and from line T.
⇒ locus of centre of C' is a parabola.
PASSAGE1
ABCD is a square of side length 2 units. C_{1 }is the circle touching all the sides of the square ABCD and C_{2} is the circumcircle of square ABCD. L is a fixed line in the same plane and R is a fixed point.
Q.3. A line L' through A is drawn parallel to BD. Point S moves such that its distances from the line BD and the vertex A are equal. If locus of S cuts L' at T_{2 }and T_{3 }and AC at T_{1}, then area of ΔT_{1}T_{2}T_{3} is (2006  5M, –2)
(a) sq. units
(b) sq. units
(c) 1 sq. units
(d) 2 sq. units
Ans. Sol. (c) Since S is equidistant form A and line BD, it traces a parabola. Clearly, AC is the axis, A (1, 1) is the focus
and is the vertex of parabola.
T_{2} T_{3 }= latus rectum of parabola
∴ Area (ΔT_{1}T_{2}T_{3}) =
PASSAGE2
A circle C of radius 1 is inscribed in an equilateral triangle PQR.
The points of contact of C with the sides PQ, QR, RP are D, E, F, respectively. The line PQ is given by the equationand the point D is
Further, it is given that the origin and the centre of C are on the same side of the line PQ.
Q. 4. The equation of circle C is (2008)
(a)
(b)
(c)
(d)
Ans. Sol. (d) Slope of CD =
∴ Parametric equation of CD is
∴ Two possible coordinates of C are
or
As (0, 0) and C lie on the same side of PQ
should be the coordinates of C.
NOTE THIS STEP : Remember (x_{1}, y_{1}) and (x_{2}, y_{2}) lie on the same or opposite side of a line ax + by + c = 0 according as
∴ Equation of the circle is
PASSAGE2
A circle C of radius 1 is inscribed in an equilateral triangle PQR.
The points of contact of C with the sides PQ, QR, RP are D, E, F, respectively. The line PQ is given by the equationand the point D is
Further, it is given that the origin and the centre of C are on the same side of the line PQ.
Q. 5. Points E and F are given by (2008)
(a)
(b)
(c)
(d)
Ans. Sol. (a) ΔPQR is an equilateral triangle, the incentre C must coincide with centriod of ΔPQR and D, E, F must concide with the mid points of sides PQ, QR and RP respectively.
Also
Writing the equation of side PQ in symmetric form we
get,
∴ Coordinates of P =
and
coordinates of
Let coordinates of R be (α,β) , then using the formula for centriod of Δ we get
⇒ α = 0 and β = 0
∴ Coordinates of R = (0, 0)
Now coordinates of E = mid point of QR =
and coordinates of F = mid point of PR =
PASSAGE2
A circle C of radius 1 is inscribed in an equilateral triangle PQR.
The points of contact of C with the sides PQ, QR, RP are D, E, F, respectively. The line PQ is given by the equationand the point D is
Further, it is given that the origin and the centre of C are on the same side of the line PQ.
Q. 6. Equations of the sides QR, RP are (2008)
(a)
(b)
(c)
(d)
Ans. Sol. (d) Equation of side QR is y = and equation of side RP is y = 0
Paragraph 3 Given the implicit function y3 – 3y + x = 0
For x ∈(–∞, –2) ∪ (2,∞) it is y = f (x) real valued differentiable function and for x ∈ (–2, 2) it is y = g(x) real valued differentiable function.
PASSAGE3
A tangent PT is drawn to the circle x^{2} + y^{2} = 4 at the point . A straight line L, perpendicular to PT is a tangent to the circle (x – 3)^{2} + y^{2} = 1. (2012)
Q. 7. A possible equation of L is
(a)
(b)
(c)
(d)
Ans. Sol. (a) Equation of tangent PT to the circle x^{2} + y^{2} = 4 at the point
Let the line L, perpendicular to tangent PT be
As it is tangent to the circle (x – 3)^{2} + y^{2} = 1
∴ length of perpendicular from centre of circle to the tangent = radius of circle.
= – 1 or – 5
∴ Equation of L can be
PASSAGE3
A tangent PT is drawn to the circle x^{2} + y^{2} = 4 at the point . A straight line L, perpendicular to PT is a tangent to the circle (x – 3)^{2} + y^{2} = 1. (2012)
Q. 8. A common tangent of the two circles is
(a) x = 4
(b) y = 2
(c)
(d)
Ans. d
Sol.
From the figure it is clear that the intersection point of two direct common tangents lies on xaxis.
Also ΔPT_{1}C_{1 }~ ΔPT_{2}C_{2} ⇒ PC_{1} : PC_{2 }= 2 : 1
or P divides C_{1}C_{2} in the ratio 2 : 1 externally
∴ Coordinates of P are (6, 0) Let the equation of tangent through P be y = m (x – 6)
As it touches x^{2} + y^{2} = 4
36 m^{2} = 4(m^{2 }+ 1)
∴ Equations of common tangents are
Also x = 2 is the common tangent to the two circles.
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