Question

The number of circles passing through the origin and touching the lines \(x+y = 1\) and \(x-y = 1\) is \(\ldots\)

• A. \(1\)
• B. \(2\)
• C. \(3\)
• D. \(4\)

Hint:

The centre lies on an angle bisector of the two lines. Check each bisector separately.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
A circle touching two lines has its centre at equal distances from both, so its centre lies on one of the angle bisectors. The lines \(x + y = 1\) and \(x - y = 1\) meet at \((1, 0)\) and are perpendicular. Their bisectors are \(y = 0\) and \(x = 1\).

Step 2: Key Formula or Approach:
For a circle with centre \(C\) passing through the origin, the radius equals \(|OC|\). It also equals the distance from \(C\) to each line.

Step 3: Case 1: centre on y = 0:
Let the centre be \((h, 0)\). Radius \(= |h|\).
Distance from \((h,0)\) to \(x + y - 1 = 0\) is \(\dfrac{|h - 1|}{\sqrt{2}}\).
\[ \frac{|h-1|}{\sqrt{2}} = |h| \Rightarrow (h-1)^2 = 2h^2 \Rightarrow h^2 + 2h - 1 = 0 \]
The discriminant is \(4 + 4 = 8 > 0\), so there are two real values \(h = -1 \pm \sqrt{2}\). That gives 2 circles.

Step 4: Case 2: centre on x = 1:
Let the centre be \((1, k)\). Radius squared \(= 1 + k^2\). Distance to \(x + y = 1\) is \(\dfrac{|k|}{\sqrt{2}}\).
\[ \frac{k^2}{2} = 1 + k^2 \Rightarrow k^2 = -2 \]
This has no real solution. So no circle exists here.

Final Answer:
Only case 1 gives circles, and it gives 2, so the answer is option (B). \[ \boxed{2 \text{ (B)}} \]