Question

The diameters of a circle are along $2x+y-7=0$ and $x+3y-11=0$. Then, the equation of this circle, which also passes through (5, 7), is

• A. $x^{2}+y^{2}-4x-6y-16=0$
• B. $x^{2}+y^{2}-4x-6y-20=0$
• C. $x^{2}+y^{2}-4x-6y-12=0$
• D. $x^{2}+y^{2}+4x+6y-12=0$

Hint:

The center of a circle is the intersection of any two of its diameters.

Answer 1

The Correct Option is C

Step 1: Find the Center
The intersection point of the diameters is the center of the circle. Solving the equations $2x+y=7$ and $x+3y=11$: Multiply the second equation by 2: $2x+6y=22$. Subtract the first from this: $(2x+6y) - (2x+y) = 22 - 7 \Rightarrow 5y = 15 \Rightarrow y = 3$. Substitute $y=3$ into $2x+y=7$: $2x+3=7 \Rightarrow 2x=4 \Rightarrow x=2$. The center is $(2, 3)$.
Step 2: Find the Radius
The circle passes through $(5, 7)$. Radius $R = \sqrt{(5-2)^{2}+(7-3)^{2}} = \sqrt{3^{2}+4^{2}} = \sqrt{9+16} = 5$.
Step 3: Determine the Equation
The equation is $(x-2)^{2}+(y-3)^{2}=5^{2}$. $x^{2}-4x+4 + y^{2}-6y+9 = 25$ $x^{2}+y^{2}-4x-6y-12=0$.
Final Answer: (c)