Question

The radius of the circle having three chords along y-axis, the line $y=x$ and the line $2x+3y=10$ is

• A. $\frac{10}{\sqrt{13}}$
• B. $\frac{\sqrt{26}}{3}$
• C. $\frac{5}{\sqrt{13}}$
• D. $\frac{10}{3}$

Hint:

The circumcircle of a triangle is the unique circle that passes through all three of its vertices. If one of the vertices is at the origin, the equation of the circumcircle takes the simpler form $x^2+y^2+2gx+2fy=0$, as the constant term 'c' is zero.

Answer 1

The Correct Option is B

Step 1: Understand the geometric configuration.
If three lines are chords of a circle, their points of intersection must lie on that circle. Therefore, the circle in question is the circumcircle of the triangle formed by the intersection of the three given lines.

Step 2: Find the vertices of the triangle.
The three lines are $L_1: x=0$ (y-axis), $L_2: y=x$, and $L_3: 2x+3y=10$. Vertex A (intersection of $L_1$ and $L_2$): Substitute $x=0$ into $y=x$, gives $y=0$. So, $A=(0,0)$. Vertex B (intersection of $L_1$ and $L_3$): Substitute $x=0$ into $2x+3y=10$, gives $3y=10 \implies y=10/3$. So, $B=(0, 10/3)$. Vertex C (intersection of $L_2$ and $L_3$): Substitute $y=x$ into $2x+3y=10$, gives $2x+3x=10 \implies 5x=10 \implies x=2$. Since $y=x$, $y=2$. So, $C=(2,2)$.

Step 3: Find the equation of the circumcircle.
Since the circle passes through the origin $A(0,0)$, its equation is of the form $x^2+y^2+2gx+2fy=0$. The points B and C must satisfy this equation. Substitute B(0, 10/3): $0^2 + (10/3)^2 + 2g(0) + 2f(10/3) = 0 \implies \frac{100}{9} + \frac{20f}{3} = 0$. $\frac{20f}{3} = -\frac{100}{9} \implies f = -\frac{100}{9} \cdot \frac{3}{20} = -\frac{5}{3}$. Substitute C(2,2): $2^2 + 2^2 + 2g(2) + 2f(2) = 0 \implies 8 + 4g + 4f = 0 \implies 2+g+f=0$. Substitute $f=-5/3$: $2+g-\frac{5}{3}=0 \implies g = \frac{5}{3}-2 = -\frac{1}{3}$.

Step 4: Calculate the radius of the circle.
The center of the circle is $(-g, -f) = (1/3, 5/3)$. The radius $R$ is the distance from the center to any vertex. Using the origin is easiest. \[ R = \sqrt{(1/3 - 0)^2 + (5/3 - 0)^2} = \sqrt{\frac{1}{9} + \frac{25}{9}} = \sqrt{\frac{26}{9}}. \] \[ \boxed{R = \frac{\sqrt{26}}{3}}. \]