IGDTUW 2026 — Mathematics PYQ

Step 1: Write down the equations of the given circles

• Let the first circle be:

  $$S_1\equiv x^2+y^2+4x+22y+c=0\text{— (Equation 1)}$$

• Let the second circle be:

  $$S_2\equiv x^2+y^2-2x+8y-d=0\text{— (Equation 2)}$$

Step 2: Find the center of the second circle ($S_2=0$)

The standard equation of a circle is given by $x^2+y^2+2gx+2fy+c^{\prime }=0$, where its center is $(-g,-f)$.

Comparing this standard form with $S_2$:

• $2g=-2⟹g=-1$

• $2f=8⟹f=4$

Thus, the center of the second circle $S_2$ is:

$$\text{Center }(C_2)=(-g,-f)=(1,-4)$$

Step 3: Find the equation of the common chord

The equation of the common chord of two intersecting circles $S_1=0$ and $S_2=0$ is obtained by subtracting one equation from the other:

$$S_1-S_2=0$$

Let's compute $(x^2+y^2+4x+22y+c)-(x^2+y^2-2x+8y-d)=0$:

$$(4x-(-2x))+(22y-8y)+(c-(-d))=0$$

$$6x+14y+c+d=0\text{— (Equation 3)}$$

Step 4: Substitute the center into the common chord equation

Since circle $S_1$ bisects the circumference of $S_2$, this common chord (Equation 3) must pass through the center of $S_2$, which is $(1,-4)$.

Substituting $x=1$ and $y=-4$ into Equation 3:

$$6(1)+14(-4)+c+d=0$$

$$6-56+c+d=0$$

$$-50+c+d=0$$

$$c+d=50$$

Correct Answer

(b) $50$