Question

The radius of a circle $C_1$ is thrice the radius of another circle $C_2$ and the centres of $C_1$ and $C_2$ are (1,2) and (3,-2) respectively. If they cut each other orthogonally and the radius of the circle $C_1$ is 3r, then the equation of the circle with r as radius and (1,-2) as centre is

• A. $x^2+y^2-2x+4y-3=0$
• B. $x^2+y^2-2x+4y+7=0$
• C. $x^2+y^2-2x+4y-7=0$
• D. $x^2+y^2-2x+4y+3=0$

Hint:

The condition for orthogonality, $d^2 = r_1^2 + r_2^2$, is fundamental. It arises from applying the Pythagorean theorem to the triangle formed by the two centers and one of their intersection points, where the tangents at that point are perpendicular.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
This problem involves the condition for two circles to intersect orthogonally. We are given information about their radii and centers, which allows us to find the value of an unknown parameter $r$. Then, we use this value of $r$ to write the equation of a third circle.
Step 2: Key Formula or Approach
1. Let the radii of circles $C_1$ and $C_2$ be $r_1$ and $r_2$, and their centers be $O_1$ and $O_2$. 2. The condition for two circles to cut orthogonally is that the square of the distance between their centers is equal to the sum of the squares of their radii: $d^2 = r_1^2 + r_2^2$. 3. Use the given information to set up an equation and solve for $r$. 4. Write the equation of the required circle using the standard form $(x-h)^2+(y-k)^2=r^2$.
Step 3: Detailed Explanation
1. Identify the given information: Center of $C_1$: $O_1 = (1,2)$. Radius of $C_1$: $r_1 = 3r$. Center of $C_2$: $O_2 = (3,-2)$. Radius of $C_2$: $r_2$. We are given that the radius of $C_1$ is thrice the radius of $C_2$. So, $r_1 = 3r_2$. Since $r_1=3r$, this implies $3r = 3r_2$, so $r_2=r$. 2. Apply the orthogonality condition: First, calculate the square of the distance, $d^2$, between the centers $O_1(1,2)$ and $O_2(3,-2)$. \[ d^2 = (3-1)^2 + (-2-2)^2 = (2)^2 + (-4)^2 = 4 + 16 = 20 \] The orthogonality condition is $d^2 = r_1^2 + r_2^2$. Substitute the known values: \[ 20 = (3r)^2 + (r)^2 \] \[ 20 = 9r^2 + r^2 \] \[ 20 = 10r^2 \] \[ r^2 = 2 \] So, the radius of the third circle is $r = \sqrt{2}$. 3. Find the equation of the required circle: We need the equation of a circle with radius $r = \sqrt{2}$ (so $r^2=2$) and center $(h,k)=(1,-2)$. The equation is $(x-h)^2 + (y-k)^2 = r^2$. \[ (x-1)^2 + (y-(-2))^2 = 2 \] \[ (x-1)^2 + (y+2)^2 = 2 \] Expand the equation: \[ (x^2 - 2x + 1) + (y^2 + 4y + 4) = 2 \] \[ x^2 + y^2 - 2x + 4y + 5 = 2 \] \[ x^2 + y^2 - 2x + 4y + 3 = 0 \] Step 4: Final Answer
The equation of the required circle is $x^2+y^2-2x+4y+3=0$.