NIMCET 2013 — Mathematics PYQ

Step 1: General Equation of a Circle

The general equation of a circle is:

The centre of this circle is $(-g,-f)$.

Step 2: Use the Condition for the Centre

The centre $(-g,-f)$ lies on the line $x+2y+3=0$. Substituting the centre into the line equation:

Step 3: Pass the Circle Through Given Points

The circle passes through $(-1,1)$ and $(2,1)$.

• For $(-1,1)$:

  $$(-1{)}^2+(1{)}^2+2g(-1)+2f(1)+c=0$$
  $$1+1-2g+2f+c=0⟹-2g+2f+c=-2\text{— (Equation 2)}$$

• For $(2,1)$:

  $$(2{)}^2+(1{)}^2+2g(2)+2f(1)+c=0$$
  $$4+1+4g+2f+c=0⟹4g+2f+c=-5\text{— (Equation 3)}$$

Step 4: Solve the Equations

Subtract (Equation 2) from (Equation 3) to eliminate $f$ and $c$:

Substitute $g=-\frac{1}{2}$ into (Equation 1):

Substitute $g$ and $f$ into (Equation 2) to find $c$:

Step 5: Form the Final Equation

Substitute $g=-\frac{1}{2}$, $f=\frac{7}{4}$, and $c=-\frac{13}{2}$ into the general equation:

Multiply the entire equation by 2 to clear the denominators:

Final Answer: The equation is $2x^2+2y^2-2x+7y-13=0$.