Question

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

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

Answer 1

The correct option is C

${\mathrm{x}}^2+{\mathrm{y}}^2–4\mathrm{x}–6\mathrm{y}-12=0$

Explanation for correct option:

Finding the equation of the circle.

Given, the diameters of the circle are $2\mathrm{x}+\mathrm{y}=7$ and $\mathrm{x}+3\mathrm{y}=11$ , solving them gives us centre $=(2,3)$

The circle passes through the point $(5,7)$.

By solving the two diameters we will get the coordinates of the center of the circle.

Equating the two diameters we get,

$2\mathrm{x}+\mathrm{y}=7$and $\mathrm{x}+3\mathrm{y}=11$,Center $=(2,3)$.

Radius of the circle is the distance between the center and any other point on the circle. Therefore, the radius of the circle will be equal to the distance between $(2,3)$and $(5,7)$.
$$\begin{gathered} =\sqrt{(5-2{)}^2+(7-3{)}^2} \\ =\sqrt{3^2+4^2}-5 \end{gathered}$$

Equation of circle will be,
$$\begin{gathered} \Rightarrow (x-2{)}^2+(y-3{)}^2=5^2 \\ \Rightarrow x^2-4x+4+y^2-6y+9=25 \\ \Rightarrow x^2-4x+y^2-6y-12=0 \\ \Rightarrow x^2+y^2-4x-6y-12=0 \end{gathered}$$

Hence, the correct answer is Option (C).