Question:
Number of circles intersecting
\[
x^2+y^2=4,\quad x^2+y^2-2x-3=0
\]
and
\[
x^2+y^2-2y-3=0
\]
orthogonally is:
Number of circles intersecting
\[
x^2+y^2=4,\quad x^2+y^2-2x-3=0
\]
and
\[
x^2+y^2-2y-3=0
\]
orthogonally is:
Show Hint
For two circles to cut orthogonally, use \(2gg_1+2ff_1=c+c_1\). After finding the circle, always check whether its radius squared is non-negative.
Updated On: Jun 18, 2026
- \(0\)
- \(1\)
- \(2\)
- \(\infty\)
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The Correct Option is A
Solution and Explanation
Step 1: Consider a general circle.
Let the required circle be \[ x^2+y^2+2gx+2fy+c=0. \] For two circles \[ x^2+y^2+2gx+2fy+c=0 \] and \[ x^2+y^2+2g_1x+2f_1y+c_1=0 \] to intersect orthogonally, the condition is \[ 2gg_1+2ff_1=c+c_1. \]
Step 2: Apply orthogonality with \(x^2+y^2=4\).
Here, \[ x^2+y^2-4=0. \] So, \[ g_1=0,\quad f_1=0,\quad c_1=-4. \] Using the orthogonality condition, \[ 0=c-4. \] Thus, \[ c=4. \]
Step 3: Apply orthogonality with \(x^2+y^2-2x-3=0\).
Here, \[ g_1=-1,\quad f_1=0,\quad c_1=-3. \] So, \[ 2g(-1)+2f(0)=c-3. \] \[ -2g=4-3. \] \[ -2g=1. \] \[ g=-\frac12. \]
Step 4: Apply orthogonality with \(x^2+y^2-2y-3=0\).
Here, \[ g_1=0,\quad f_1=-1,\quad c_1=-3. \] So, \[ 2g(0)+2f(-1)=c-3. \] \[ -2f=4-3. \] \[ -2f=1. \] \[ f=-\frac12. \]
Step 5: Check whether the obtained circle is real.
The required circle becomes \[ x^2+y^2-x-y+4=0. \] For a circle \[ x^2+y^2+2gx+2fy+c=0, \] radius squared is \[ r^2=g^2+f^2-c. \] Here, \[ g=-\frac12,\quad f=-\frac12,\quad c=4. \] Thus, \[ r^2=\frac14+\frac14-4. \] \[ r^2=\frac12-4=-\frac72. \] Since \(r^2<0\), no real circle exists.
Step 6: Final conclusion.
Therefore, the number of real circles is \[ \boxed{0} \]
Let the required circle be \[ x^2+y^2+2gx+2fy+c=0. \] For two circles \[ x^2+y^2+2gx+2fy+c=0 \] and \[ x^2+y^2+2g_1x+2f_1y+c_1=0 \] to intersect orthogonally, the condition is \[ 2gg_1+2ff_1=c+c_1. \]
Step 2: Apply orthogonality with \(x^2+y^2=4\).
Here, \[ x^2+y^2-4=0. \] So, \[ g_1=0,\quad f_1=0,\quad c_1=-4. \] Using the orthogonality condition, \[ 0=c-4. \] Thus, \[ c=4. \]
Step 3: Apply orthogonality with \(x^2+y^2-2x-3=0\).
Here, \[ g_1=-1,\quad f_1=0,\quad c_1=-3. \] So, \[ 2g(-1)+2f(0)=c-3. \] \[ -2g=4-3. \] \[ -2g=1. \] \[ g=-\frac12. \]
Step 4: Apply orthogonality with \(x^2+y^2-2y-3=0\).
Here, \[ g_1=0,\quad f_1=-1,\quad c_1=-3. \] So, \[ 2g(0)+2f(-1)=c-3. \] \[ -2f=4-3. \] \[ -2f=1. \] \[ f=-\frac12. \]
Step 5: Check whether the obtained circle is real.
The required circle becomes \[ x^2+y^2-x-y+4=0. \] For a circle \[ x^2+y^2+2gx+2fy+c=0, \] radius squared is \[ r^2=g^2+f^2-c. \] Here, \[ g=-\frac12,\quad f=-\frac12,\quad c=4. \] Thus, \[ r^2=\frac14+\frac14-4. \] \[ r^2=\frac12-4=-\frac72. \] Since \(r^2<0\), no real circle exists.
Step 6: Final conclusion.
Therefore, the number of real circles is \[ \boxed{0} \]
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