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1

### WB JEE 2009

The equation of the chord of he circle $${x^2} + {y^2} - 4x = 0$$ whose midpoint is (1, 0) is

A
y = 2
B
y = 1
C
x = 2
D
x = 1

## Explanation

Equation of chord of circle

x2 + y2 + 2gx + 2fy + c = 0, where midpoint is (x1, y1) is T = S1

where T = xx1 + yy1 + g(x + x1) + f(y + y1) + c

and S1 = x$$_1^2$$ + y$$_1^2$$ + 2gx1 + 2fy1 + c

$$\therefore$$ Equation of chord of the circle x2 + y2 $$-$$ 4x = 0 whose midpoint is (1, 0) is

1x + 0y $$-$$ 2(x + 1) + 0(y + 0) = 12 + 02 $$-$$ 4 . 1

$$\Rightarrow$$ $$-$$ x $$-$$ 2 = $$-$$ 3 $$\Rightarrow$$ x = 1

2

### WB JEE 2008

The locus of the centres of the circles which touch both the axes is given by

A
$${x^2} - {y^2} = 0$$
B
$${x^2} + {y^2} = 0$$
C
$${x^2} - {y^2} = 1$$
D
$${x^2} + {y^2} = 1$$

## Explanation

Let the coordinates of centre of circle are (h, k) and radius is a unit. Circle touches both the axes

h = a ..... (i)

k = a ..... (ii)

Eliminating a by squaring (i) and (ii) and then subtracting we get $${h^2} - {k^2} = 0$$

$$\Rightarrow {x^2} - {y^2} = 0$$, which is required equation of locus.

3

### WB JEE 2008

The circles $${x^2} + {y^2} + 6x + 6y = 0$$ and $${x^2} + {y^2} - 12x - 12y = 0$$

A
cut orthogonally
B
touch each other internally
C
intersect in two points
D
touch each other externally

## Explanation

Given equations of circles are

$${x^2} + {y^2} + 6x + 6y = 0$$ ..... (i)

$${x^2} + {y^2} - 12x - 12y = 0$$ .... (ii)

Centre $$\equiv$$ $${C_1}( - 3, - 3)$$

Centre $$\equiv$$ $${C_2}(6,6)$$

$${r_1} = \sqrt {9 + 9} = \sqrt {18} = 3\sqrt 2$$

$${r_2} = \sqrt {36 + 36} = 6\sqrt 2$$

$${C_1}{C_2} = \sqrt {{{(6 + 3)}^2} + {{(6 + 3)}^2}} = \sqrt {81 + 81} = 9\sqrt 2$$

Since, $${C_1}{C_2} = {r_1} + {r_2} = 9\sqrt 2$$

$$\therefore$$ both circles touch each other externally.

4

### WB JEE 2008

The equation $$(x - {x_1})(x - {x_2}) + (y - {y_1})(y - {y_2}) = 0$$ represents a circle whose centre is

A
$$\left( {{{{x_1} - {x_2}} \over 2},{{{y_1} - {y_2}} \over 2}} \right)$$
B
$$\left( {{{{x_1} + {x_2}} \over 2},{{{y_1} + {y_2}} \over 2}} \right)$$
C
$$({x_1},{y_1})$$
D
$$({x_2},{y_2})$$

## Explanation

$$(x - {x_1})(x - {x_2}) + (y - {y_1})(y - {y_2}) = 0$$

This is equation of circle in diameter form where $$({x_1},{y_1})$$ and $$({x_2},{y_2})$$ are the end points of the diameter.

$$\therefore$$ Centre is midpoint of diameter $$= \left( {{{{x_1} + {x_2}} \over 2},{{{y_1} + {y_2}} \over 2}} \right)$$.

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