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: Understanding the Concept
We need to find the center of circle C using the given geometric conditions. Then, we need to check which of the given circle equations is satisfied by the coordinates of this center.
Step 2: Key Formula or Approach
1. Let the center of circle C be $(h,k)$ and its radius be $r$. 2. Condition 1: Circle touches the X-axis. This means the radius is equal to the absolute value of the y-coordinate of the center, i.e., $r = |k|$. The equation is $(x-h)^2+(y-k)^2 = k^2$. 3. Condition 2: Intercept on the Y-axis is 2. The length of the intercept made by a circle $(x-h)^2+(y-k)^2=r^2$ on the Y-axis ($x=0$) is $2\sqrt{r^2-h^2}$. 4. Condition 3: The center $(h,k)$ lies on the line $y=x+1$, so $k=h+1$. 5. Solve these equations to find $(h,k)$. 6. Substitute $(h,k)$ into the option equations to see which one holds true.
Step 3: Detailed Explanation
1. Use the intercept condition: The radius is $r=|k|$. The Y-intercept length is $2\sqrt{r^2-h^2} = 2$. \[ \sqrt{k^2 - h^2} = 1 \] \[ k^2 - h^2 = 1 \] 2. Use the line condition: The center $(h,k)$ lies on $y=x+1$, so we have: \[ k = h+1 \] 3. Solve for h and k: Substitute the expression for $k$ from the line condition into the equation from the intercept condition: \[ (h+1)^2 - h^2 = 1 \] \[ (h^2+2h+1) - h^2 = 1 \] \[ 2h+1 = 1 \implies 2h=0 \implies h=0 \] Now find $k$: \[ k = h+1 = 0+1 = 1 \] So, the center of the circle C is the point $(0,1)$. 4. Check the options: We need to find which of the given circles passes through the point $(0,1)$. We substitute $x=0, y=1$ into each equation. (A) $0^2+1^2-2(0)-4(1)+1 = 1-4+1 = -2 \neq 0$. (B) $0^2+1^2-26(0)-20(1)+19 = 1-20+19 = 0$. This is satisfied. (C) $0^2+1^2-20(0)-26(1)+19 = 1-26+19 = -6 \neq 0$. (D) $0^2+1^2+2(0)-4(1)+1 = 1-4+1 = -2 \neq 0$. Step 4: Final Answer
The center of circle C is $(0,1)$. The only circle from the options that passes through this point is $x^2+y^2-26x-20y+19=0$.