Question

Equation of a circle whose area is 154 sq units and having \(2x-3y+12=0\) and \(x+4y-5=0\) as diameters is

• A. \(x^2+y^2+6x-4y+36=0\)
• B. \(x^2-y^2+6x-4y-36=0\)
• C. \(x^2+y^2-6x+4y-36=0\)
• D. \(x^2+y^2+6x-4y-36=0\)

Hint:

If two diameters of a circle are given as lines, their point of intersection gives the centre of the circle.

Answer 1

The Correct Option is D

Step 1: Understand the given information.
The circle has area \(154\) sq units and the given two lines are diameters of the circle.
\[ 2x-3y+12=0 \] and
\[ x+4y-5=0. \]

Step 2: Find the radius using area.
Area of a circle is:
\[ \pi r^2 = 154. \]
Taking \(\pi=\frac{22}{7}\), we get:
\[ \frac{22}{7}r^2=154. \]
\[ r^2=154 \times \frac{7}{22}=49. \]
So,
\[ r=7. \]

Step 3: Find the centre of the circle.
Since both given lines are diameters, their point of intersection is the centre of the circle.
Solve:
\[ 2x-3y+12=0 \] and
\[ x+4y-5=0. \]

Step 4: Solve the equations.
From the second equation:
\[ x=5-4y. \]
Substitute in the first equation:
\[ 2(5-4y)-3y+12=0. \]
\[ 10-8y-3y+12=0. \]
\[ 22-11y=0. \]
\[ y=2. \]
Now,
\[ x=5-4(2)=-3. \]
So, the centre is:
\[ (-3,2). \]

Step 5: Use standard equation of circle.
The equation of circle with centre \((h,k)\) and radius \(r\) is:
\[ (x-h)^2+(y-k)^2=r^2. \]
Here,
\[ h=-3,\quad k=2,\quad r=7. \]
Therefore:
\[ (x+3)^2+(y-2)^2=49. \]

Step 6: Expand the equation.
\[ x^2+6x+9+y^2-4y+4=49. \]
\[ x^2+y^2+6x-4y+13=49. \]

Step 7: Bring all terms to one side.
\[ x^2+y^2+6x-4y-36=0. \]
Final Answer:
\[ \boxed{x^2+y^2+6x-4y-36=0}. \]