## A(2, 0), B(0, 2) and C are the vertices of a triangle inscribed in the circle $x^2 + y^2 = 4$ …

Problem:   A(2, 0), B(0, 2) and C are the vertices of a triangle inscribed in the circle $\large x^2 + y^2 = 4$ . The bisector of angle C belongs to either of two families of concurrent lines, whose points of concurrency are

(A) (√2 , √2 )

(B) (-√2 , -√2 )

(C) (√2 , -√2 )

(D) (-√2 , √2 )

Ans:    (A), (B).

Sol: The points of concurrency lie on the intersection of perpendicular bisectors of AB with the circle x2 + y2 = 4.

Hence the points are (√2 , √2 )   or (-√2 , -√2 )

## $\frac{x}{a} + \frac{y}{b} = 1$ and $\frac{x}{c} + \frac{y}{d} = 1$ intersect the axes at four concyclic points…

Problem: $\large \frac{x}{a} + \frac{y}{b} = 1$ and $\large \frac{x}{c} + \frac{y}{d} = 1$ intersect the axes at four concyclic points and c2 + a2 = b2 + d2, then these lines can intersect at (a, b, c, d > 0)

(A) (1, 1)

(B) (1, – 1)

(C) (2, – 2)

(D) (3, – 2)

Ans: (A) , (B), (C)

Sol: Points A, B, C, D are concyclic, then ac = bd.

The co-ordinate of the points of intersection of lines are

$\large (\frac{ac(b-d)}{bc-ad}) , (\frac{bd(c-a)}{bc-ad})$

given c2 + a2 = b2 + d2    ( ac = bd)

⇒ (c – a)2 = (b – d)2

⇒ c – a = ± (b – d)

then the locus of the points of intersection is y = ± x.

## If x + y = 0 is the angle bisector of the angle containing the point (1, 0), for the line  3x + 4y + b = 0, 4x + 3y – b = 0 then b can be

Problem:   If x + y = 0 is the angle bisector of the angle containing the point (1, 0), for the line  3x + 4y + b = 0, 4x + 3y – b = 0 then b can be

(A) 3

(B) 4

(C) 5

(D) 6

Ans: (C) , (D)

Sol: (3x + 4y + b) = ± (4x + 3y – b)

For x + y = 0, we have to choose -ve sign.

(3x1 + 4y1 + b) (4x1 + 3y1 – b) < 0

⇒ (3 + b) (4 – b) < 0

⇒ b > 4 or b < -3.

## ABCD is a square. E(4, 3), F(2, 5) lie on AB and CD respectively such that EF divides the square in two equal parts…

Problem: ABCD is a square. E(4, 3), F(2, 5) lie on AB and CD respectively such that EF divides the square in two equal parts. If the coordinates of A is (7, 3), other coordinates of the vertices can be

(A) (7, 2)

(B) (7, 5)

(C) (-1, 3)

(D) (-1, 7)

Ans: (B), (C).

Sol: Mid-point (G) of EF is centre of square.

## The minimum distance of $4x^2 + y^2 + 4x – 4y + 5 = 0$ from the line $-4x + 3 y = 3$ is

Problem: The minimum distance of $\large 4x^2 + y^2 + 4x – 4y + 5 = 0$ from the line $-4x + 3 y = 3$ is

(A) 2

(B) 3

(C) 1

(D) none of these

Ans: (C)

Sol: The given curve represents the point (-1/2 , 2) , so minimum distance = 1.

## Through the point P(α , β), where α β > 0, the straight line $\large \frac{x}{a} + \frac{y}{b} = 1$ is

Problem:     Through the point P(α , β), where α β > 0, the straight line $\large \frac{x}{a} + \frac{y}{b} = 1$ is drawn so as to form with coordinate axes a triangle of area s. If ab > 0, then least value of s is

(A) 2α β

(B) (1/2 ) α β

(C) α β

(D) none of these

Ans: (A)

Sol: Given P ≡ (α , β)

Given line is $\large \frac{x}{a} + \frac{y}{b} = 1$  …(i)

If line (i) cuts x and y axes at A and B respectively, then

A ≡ (a, 0) and B ≡ (0, b).

Also the area of ΔOAB = s

(1/2)ab = s ⇒  ab = 2s

Since line (i) passes through P(α , β),

$\large \frac{\alpha}{a} + \frac{\beta}{b} = 1$

$\large \frac{\alpha}{a} + \frac{\alpha \beta}{2 s} = 1$

⇒ a2 β – 2a s + 2α s = 0

Since a is real, 4s2 – 8 α β s  ≥ 0

⇒  s ≥ 2 α β  .

Hence the least value of  s = 2 α β

## If the lines x=a + m, y=-2 and y = mx are concurrent, the least value of  |a| is

Problem:  If the lines x=a + m, y=-2 and y = mx are concurrent, the least value of  |a| is

(A) 0

(B) √2

(C) 2√2

(D) none of these

Ans:(C)

Sol: Since the lines are concurrent

$\large \left| \begin{array}{ccc} 1 & 0 & -a-m \\ 0 & 1 & 2 \\ m & -1 & 0 \end{array} \right| = 0$

⇒ m2 + am + 2 = 0

Since m is real, a2  ≥ 8,  |a| ≥ 2√2

Hence the least value of |a| is 2√2.