Question

A circle C touches the X-axis and makes an intercept of length 2 units on the Y-axis. If the centre of this circle lies on the line $y=x+1$, then a circle passing through the centre of the circle C is

• A. $x^2+y^2-2x-4y+1=0$
• B. $x^2+y^2-26x-20y+19=0$
• C. $x^2+y^2-20x-26y+19=0$
• D. $x^2+y^2+2x-4y+1=0$

Hint:

Remember these key properties for circles with center $(h,k)$ and radius $r$: - Touches X-axis: $r=|k|$. - Touches Y-axis: $r=|h|$. - X-intercept length: $2\sqrt{r^2-k^2}$. - Y-intercept length: $2\sqrt{r^2-h^2}$. These shortcuts can save a lot of time.

Answer 1

The Correct Option is B

Step 1: Find Centre of Circle C: Let centre be \( (h,k) \). - Touches X-axis \( \implies \) Radius \( r = |k| \). - Intercept on Y-axis = 2 \( \implies 2\sqrt{r^2 - h^2} = 2 \implies r^2 - h^2 = 1 \). - Centre lies on \( y = x+1 \implies k = h+1 \). Substitute \( r^2 = k^2 = (h+1)^2 \) into intercept equation: \( (h+1)^2 - h^2 = 1 \) \( 2h + 1 = 1 \implies h = 0 \). Then \( k = 1 \). Centre of C is \( (0,1) \).
Step 2: Check Options: The question asks for a circle passing through the centre of C, i.e., point \( (0,1) \). Substitute \( (0,1) \) into options: (A) \( 0+1-0-4+1 = -2 \neq 0 \) (B) \( 0+1-0-20+19 = 0 \) (Correct) (C) \( 0+1-0-26+19 = -6 \neq 0 \) (D) \( 0+1+0-4+1 = -2 \neq 0 \)