Question

If two circles \(2x^2+2y^2-3x+6y+k=0\) and \(x^2+y^2-4x+10y+16=0\) cut orthogonally, then the value of \(k\) is

• A. 41
• B. 14
• C. 4
• D. 1

Hint:

For circles \(x^2+y^2+2g_1x+2f_1y+c_1=0\) and \(x^2+y^2+2g_2x+2f_2y+c_2=0\), orthogonality condition is \(2(g_1g_2+f_1f_2)=c_1+c_2\).

Answer 1

The Correct Option is C

Step 1: Write circles in standard form.
Circle 1: divide by 2:
\[ x^2+y^2-\frac{3}{2}x+3y+\frac{k}{2}=0 \] So:
\[ 2g_1=-\frac{3}{2}\Rightarrow g_1=-\frac{3}{4},\quad 2f_1=3\Rightarrow f_1=\frac{3}{2},\quad c_1=\frac{k}{2} \] Circle 2:
\[ x^2+y^2-4x+10y+16=0 \] So:
\[ 2g_2=-4\Rightarrow g_2=-2,\quad 2f_2=10\Rightarrow f_2=5,\quad c_2=16 \] Step 2: Condition for orthogonality.
Two circles cut orthogonally if:
\[ 2(g_1g_2+f_1f_2)=c_1+c_2 \] Step 3: Substitute values.
\[ 2\left[\left(-\frac{3}{4}\right)(-2)+\left(\frac{3}{2}\right)(5)\right] =\frac{k}{2}+16 \] Compute inside:
\[ \left(-\frac{3}{4}\right)(-2)=\frac{3}{2},\quad \left(\frac{3}{2}\right)(5)=\frac{15}{2} \] Sum:
\[ \frac{3}{2}+\frac{15}{2}=\frac{18}{2}=9 \] So LHS:
\[ 2(9)=18 \] Thus:
\[ 18=\frac{k}{2}+16 \Rightarrow \frac{k}{2}=2 \Rightarrow k=4 \] Final Answer: \[ \boxed{4} \]