NIMCET 2007 — Mathematics PYQ

To find the equation of a circle, we need two main components:

1. The coordinates of the center of the circle $(h,k)$.

2. The radius of the circle $r$.

The standard equation of a circle is given by:

$$(x-h{)}^2+(y-k{)}^2=r^2$$

Step 1: Find the Center of the Circle

The problem states that the two given lines lie along the diameters of the circle. Since all diameters of a circle intersect exactly at its center, solving these two linear equations simultaneously will give us the coordinates of the center $(h,k)$.

The given line equations are:

1. $2x+3y+1=0⟹2x+3y=-1\text{— (Equation 1)}$

2. $3x-y-4=0⟹3x-y=4\text{— (Equation 2)}$

Let's multiply Equation 2 by $3$ to eliminate $y$:

$$3\times (3x-y)=3\times 4$$

$$9x-3y=12\text{— (Equation 3)}$$

Now, add Equation 1 and Equation 3:

$$(2x+3y)+(9x-3y)=-1+12$$

$$11x=11$$

$$x=1$$

Substitute $x=1$ back into Equation 2 to find $y$:

$$3(1)-y=4$$

$$3-y=4⟹y=3-4=-1$$

Thus, the center of the circle is $(h,k)=(1,-1)$.

Step 2: Find the Radius of the Circle

We are given that the circumference of the circle is $10\pi$. The formula for the circumference of a circle is $2\pi r$.

$$2\pi r=10\pi$$

$$2r=10⟹r=5$$

Step 3: Write the Equation of the Circle

Now substitute the center $(1,-1)$ and radius $r=5$ into the standard equation:

$$(x-1{)}^2+(y-(-1){)}^2=5^2$$

$$(x-1{)}^2+(y+1{)}^2=25$$

Expand the squared terms using the algebraic identity formulas:

$$(x^2-2x+1)+(y^2+2y+1)=25$$

$$x^2+y^2-2x+2y+2=25$$

Bring $25$ to the left-hand side to format it like the options:

$$x^2+y^2-2x+2y+2-25=0$$

$$x^2+y^2-2x+2y-23=0$$

Final Answer

The correct option is (c) $x^2+y^2-2x+2y-23=0$.