Question

Find the equation of the circle passing through the origin and the points where the line $3x+4y=12$ meets the axes of coordinates.

• A. $x^2+y^2+3x+4y=0$
• B. $x^2+y^2+3x-4y=0$
• C. $x^2+y^2-3x+4y=0$
• D. $x^2+y^2-4x-3y=0$

Answer 1

The correct option is D

$x^2+y^2-4x-3y=0$

Explanation for the correct option:

Step:1 Find out the points at which the circle cuts X and Y axis.

The equation of the given line is $3x+4y=12$.

Let, $x=0$.

Therefore, $$\begin{gathered} 3\left(0\right)+4y=12 \\ y=3 \end{gathered}$$

Therefore, the point on the Y-axis is $(0,3)$.

Let, $y=0$.

Therefore, $$\begin{gathered} 3x+4\left(0\right)=12 \\ x=4 \end{gathered}$$

Therefore, the point on the X-axis is $(4,0)$.

Therefore, this circle passes through origin and these two points.

Hence, the line that connects these two points is the diameter of the circle.

Step:2 Find out the equation of the circle

Now, the equation of the circle that has a diameter with two points at $\left(x_{1,}{x}_2\right)$ and $(y_1,y_2)$ is

$$\begin{gathered} \Rightarrow (x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0 \\ \Rightarrow (x-0)(x-4)+(y-3)(y-0)=0 \\ \Rightarrow x^2+y^2-4x-3y=0 \end{gathered}$$

Therefore, Option(D) is the correct answer.