Q.13. Suppose the two chords, drawn from the point on the circle are bisected by the y-axis. If th

Q: What is the correct answer for JEE Main Mathematics Q13?

The correct answer is B. Sol. The given circle is $x^2 + y^2 + x - 3y = 0$. The center of the circle is $C = (-\frac{1}{2}, \frac{3}{2})$. Let $P=(1,2)$ be the given point on the circle. Let the chord be PR, where R is the other end. The chord PR is bisected by the y-axis. This means the midpoint of PR lies on the y-axis. Let the midpoint be $M=(0,k)$. Since M is the midpoint of PR, and $P=(1,2)$, let $R=(x_R, y_R)$. Then

Q: Q.13. Suppose the two chords, drawn from the point on the circle are bisected by the y-axis. If the other ends of these chords are R and S, and the mid point of the line segment RS is , then is…

The correct answer is B. Sol. The given circle is $x^2 + y^2 + x - 3y = 0$. The center of the circle is $C = (-\frac{1}{2}, \frac{3}{2})$. Let $P=(1,2)$ be the given point on the circle. Let the chord be PR, where R is the other end. The chord PR is bisected by the y-axis. This means the midpoint of PR lies on the y-axis. Let the midpoint be $M=(0,k)$. Since M is the midpoint of PR, and $P=(1,2)$, let $R=(x_R, y_R)$. Then

Question

Q.13. Suppose the two chords, drawn from the point  on the circle  are bisected by the y-axis. If the other ends of these chords are R and S, and the mid point of the line segment RS is , then  is equal to :

FirstInTest · Accepted Answer

Sol. The given circle is $x^2 + y^2 + x - 3y = 0$. The center of the circle is $C = (-\frac{1}{2}, \frac{3}{2})$. Let $P=(1,2)$ be the given point on the circle. Let the chord be PR, where R is the other end. The chord PR is bisected by the y-axis. This means the midpoint of PR lies on the y-axis. Let the midpoint be $M=(0,k)$. Since M is the midpoint of PR, and $P=(1,2)$, let $R=(x_R, y_R)$. Then $\frac{1+x_R}{2}=0 \Rightarrow x_R=-1$. And $\frac{2+y_R}{2}=k \Rightarrow y_R=2k-2$. So $R=(-1, 2k-2)$. The point R lies on the circle. So $(-1)^2 + (2k-2)^2 + (-1) - 3(2k-2) = 0$. $1 + 4k^2 - 8k + 4 - 1 - 6k + 6 = 0$. $4k^2 - 14k + 10 = 0$. $2k^2 - 7k + 5 = 0$. $(2k-5)(k-1) = 0$. So $k=1$ or $k=\frac{5}{2}$. If $k=1$, then $R=(-1, 2(1)-2) = (-1,0)$. This is one end of the chord. Let's call it $S_1$. If $k=\frac{5}{2}$, then $R=(-1, 2(\frac{5}{2})-2) = (-1,5-2) = (-1,3)$. This is the other end of the chord. Let's call it $S_2$. So the two other ends of the chords are $S_1=(-1,0)$ and $S_2=(-1,3)$. The midpoint of the line segment $S_1S_2$ is $(\alpha, \beta)$. $\alpha = \frac{-1+(-1)}{2} = -1$. $\beta = \frac{0+3}{2} = \frac{3}{2}$. We need to find $6(\alpha + \beta) = 6(-1 + \frac{3}{2}) = 6(\frac{1}{2}) = 3$. The solution in the image is consistent with this derivation.

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