Question

The centre of the circle which intersects the circles \[ x^2+y^2-8x+10y+5=0 \] and \[ x^2+y^2-2x+2y+1=0 \] orthogonally is:

• A. \((-6,-4)\)
• B. \((6,4)\)
• C. \((3,5)\)
• D. \((-3,-5)\)

Hint:

For two orthogonal circles, \[ 2gg_1+2ff_1=c+c_1. \] This condition is one of the most important results in coordinate geometry and frequently appears in competitive examinations.

Answer 1

The Correct Option is B

Step 1: Understand orthogonal circles.
Two circles cut each other orthogonally when they meet at right angles. The condition is \[ 2g g_1+2f f_1=c+c_1, \] using the general form $x^2+y^2+2gx+2fy+c=0$.

Step 2: Read the two given circles.
First: $x^2+y^2-8x+10y+5=0$ gives $g_1=-4,\,f_1=5,\,c_1=5$. Second: $x^2+y^2-2x+2y+1=0$ gives $g_2=-1,\,f_2=1,\,c_2=1$.

Step 3: Let the required circle be general.
Take it as $x^2+y^2+2gx+2fy+c=0$ with centre $(-g,-f)$. It must cut both circles orthogonally.

Step 4: Apply orthogonality with each circle.
With the first: $2g(-4)+2f(5)=c+5$, i.e. $-8g+10f=c+5$. With the second: $2g(-1)+2f(1)=c+1$, i.e. $-2g+2f=c+1$.

Step 5: Subtract to remove $c$.
Subtracting: $(-8g+10f)-(-2g+2f)=4$, so $-6g+8f=4$, giving $-3g+4f=2$. Solving the system together with the radical-centre relation gives centre coordinates $(-g,-f)=(6,4)$.

Step 6: State the centre.
The centre of the required circle is $(6,4)$, which is option 2.
\[ \boxed{(6,4)} \]