MULTI CORRECT CHOICE QUESTION:-

Q.Coordinates of the centre of a circle whose radius is 2 unit and which touches the line pair `x^2 - y^2 - 2x + 1=0`

a) `(4,0)`

b) `(1 + 2sqrt2,0)`

c) `(4,1)`

d) `(1,2sqrt2)`

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First consider the line pair with the equation

`x^2-y^2-2x+1 = 0`

The left side of the equation contains complete square trinomial `x^2-2x+1` , which equals `(x-1)^2` .

So this equation can be rewritten as `(x-1)^2 = y^2` , which corresponds to the equations of two lines: `y = x-1` and `y = -x+1` .

These lines are graphed below:

If a circle is touching these two lines, it's center has to lie on the bisector of the angle formed by the lines (90 degree angle).This means the center has to either lie on the line with the equation x = 1 (and have x-coordinate x=1) or on the x-axis (and thus have coordinate y=0).

Choices a, b and d all satisfy these requirements.

Choice d: the center is at the point `(1, 2sqrt2)` . Consider the perpendicular dropped from this point to either of the lines. This is the radius of the circle in question and it has to equal 2. Since the right triangle formed by points (1,0), `(1, 2sqrt(2))` and this perpendicular has to be isosceles (bisector and the line form 45 degree angle), the length of the hypotenuse is `sqrt(2)` times 2, which is correct. So choice d is one of the correct choices.

If the center lies on the x-axis, as in choices a and b, the x-coordinate of the center `(x_0)` has to be such that `x_0-1` is also `sqrt(2) ` times 2, the radius of the circle. So `x_0=1+2sqrt(2)` , as in choice b.

**Answer: b and d**

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