Question

If $(\alpha,\beta)$ is the centre of the circle which passes through the point (1,-1) and cuts the circles $x^2 + y^2+2x-3y-5=0$, $x^2+y^2-3x+2y+1=0$ orthogonally, then $\alpha-5\beta =$

• A. -10
• B. 5
• C. -11
• D. 10

Hint:

The condition for orthogonality of two circles, $2g_1g_2+2f_1f_2=c_1+c_2$, is fundamental. When a circle needs to satisfy multiple conditions (passing through a point, orthogonality to other circles), set up a system of linear equations for its parameters $g, f, c$.

Answer 1

The Correct Option is D

Let the equation of the required circle be $x^2+y^2+2gx+2fy+c=0$.
The centre of this circle is $(-\text{g}, -f) = (\alpha, \beta)$. So, $\alpha = -g$ and $\beta = -f$.
The circle passes through the point (1,-1). So, this point must satisfy the equation:
$1^2+(-1)^2+2g(1)+2f(-1)+c=0 \implies 1+1+2g-2f+c=0 \implies 2g-2f+c=-2$. (Eq. 1)
The condition for two circles $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$ to cut orthogonally is $2g_1g_2+2f_1f_2=c_1+c_2$.
The required circle cuts $S_1: x^2+y^2+2x-3y-5=0$ orthogonally. Here, $g_1=1, f_1=-3/2, c_1=-5$. For our circle, $g_2=g, f_2=f, c_2=c$. $2g(1)+2f(-3/2) = c + (-5) \implies 2g-3f=c-5$. (Eq. 2)
The required circle cuts $S_2: x^2+y^2-3x+2y+1=0$ orthogonally. Here, $g_1=-3/2, f_1=1, c_1=1$. $2g(-3/2)+2f(1) = c+1 \implies -3g+2f=c+1$. (Eq. 3)
We now have a system of three linear equations in $g, f, c$. From Eq. 2, $c = 2g-3f+5$. Substitute this into Eq. 1 and Eq. 3.
Into Eq. 1: $2g-2f+(2g-3f+5)=-2 \implies 4g-5f=-7$. (Eq. 4)
Into Eq. 3: $-3g+2f=(2g-3f+5)+1 \implies -3g+2f=2g-3f+6 \implies 5g-5f=-6$. (Eq. 5)
Now we solve the system for $g$ and $f$ from Eq. 4 and Eq. 5.
Subtract Eq. 5 from Eq. 4: $(4g-5f)-(5g-5f)=-7-(-6) \implies -g=-1 \implies g=1$.
Substitute $g=1$ into Eq. 4: $4(1)-5f=-7 \implies 4-5f=-7 \implies 5f=11 \implies f=11/5$.
The centre is $(\alpha, \beta) = (-g, -f) = (-1, -11/5)$.
We need to find the value of $\alpha-5\beta$.
$\alpha-5\beta = (-1) - 5(-11/5) = -1 + 11 = 10$.