Circles Question - None

Question

Coordinates of the centre of a circle, whose radius is 2 unit and which touches the line pair $x^2 - y^2 - 2x + 1 = 0$ are:

Accepted Answer

Step 1: Factorize the line pair

Given: $x^2 - y^2 - 2x + 1 = 0$

Rearranging: $x^2 - 2x + 1 - y^2 = 0$

$(x-1)^2 - y^2 = 0$

$(x-1-y)(x-1+y) = 0$

The two lines are: $x - y - 1 = 0$ and $x + y - 1 = 0$

Step 2: Find angle bisectors

The angle bisectors of two lines $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ are given by:

$$\frac{a_1x + b_1y + c_1}{\sqrt{a_1^2 + b_1^2}} = \pm \frac{a_2x + b_2y + c_2}{\sqrt{a_2^2 + b_2^2}}$$

Here: $\frac{x - y - 1}{\sqrt{2}} = \pm \frac{x + y - 1}{\sqrt{2}}$

This gives: $x - y - 1 = x + y - 1$ → $y = 0$ (bisector 1)

And: $x - y - 1 = -(x + y - 1)$ → $x = 1$ (bisector 2)

Step 3: Find center on bisector with radius 2

The center must lie on one of these bisectors. Try $y = 0$: Let center be $(h, 0)$

Distance from $(h, 0)$ to line $x - y - 1 = 0$ is: $\frac{|h - 0 - 1|}{\sqrt{2}} = \frac{|h-1|}{\sqrt{2}}$

Since radius = 2: $\frac{|h-1|}{\sqrt{2}} = 2$

$|h-1| = 2\sqrt{2}$

$h = 1 + 2\sqrt{2}$ or $h = 1 - 2\sqrt{2}$

Step 4: Verify the solution

For $(1 + 2\sqrt{2}, 0)$: Distance to $x - y - 1 = 0$ is $\frac{2\sqrt{2}}{\sqrt{2}} = 2$ ✓

Distance to $x + y - 1 = 0$ is $\frac{|1 + 2\sqrt{2} - 1|}{\sqrt{2}} = \frac{2\sqrt{2}}{\sqrt{2}} = 2$ ✓

∴ Answer: b

Question: