The angle subtended by the chord $x+y-1=0$ of the circle $x^2+y^2=1$ at the origin is
- $\cos ^{-1}\left(\frac{6}{\sqrt{34}}\right)$
- $\frac{\pi}{2}$
- $\cos ^{-1}\left(\frac{2}{\sqrt{13}}\right)$
- $\frac{\pi}{3}$
Solution
Step 1: Identify the Circle and Chord: The given circle is $x^2+y^2=1$, which has its center at the origin $(0,0)$ and a radius of 1 . The chord is given by the line equation $x+y=1$.
Step 2: Find the Points of Intersection: To find the points where the chord intersects the circle, we can substitute $y=1-x$ into the circle's equation: $x^2+(1-x)^2=1$ Expanding this: $\begin{aligned} & x^2+\left(1-2 x+x^2\right)=1 \\ & 2 x^2-2 x+1=1 \\ & 2 x^2-2 x=0 \\ & 2 x(x-1)=0 \end{aligned}$ This gives us $x=0$ or $x=1$. Substituting back to find $y$ : - For $x=0, y=1 \rightarrow$ Point $(0,1)$ - For $x=1, y=0 \rightarrow$ Point $(1,0)$ Thus, the points of intersection are $(0,1)$ and $(1,0)$.
Step 3: Find the Midpoint of the Chord: The midpoint $M$ of the chord can be calculated as: $M=\left(\frac{0+1}{2}, \frac{1+0}{2}\right)=\left(\frac{1}{2}, \frac{1}{2}\right)$
Step 4: Calculate the Perpendicular Distance from the Center to the Chord : The distance $d$ from the center $(0,0)$ to the line $x+y-1=0$ can be calculated using the formula for the distance from a point to a line: $d=\frac{\left|A x_1+B y_1+C\right|}{\sqrt{A^2+B^2}}$ Here, $A=1, B=1, C=-1$, and $\left(x_1, y_1\right)=(0,0)$ : $d=\frac{|1(0)+1(0)-1|}{\sqrt{1^2+1^2}}=\frac{|-1|}{\sqrt{2}}=\frac{1}{\sqrt{2}}$
Step 5: Use Trigonometry to Find the Angle: Let $O$ be the center of the circle, $B$ be the point $(0,1)$, and $D$ be the foot of the perpendicular from $O$ to the chord. In triangle $O D B$ : $O B=1 \quad$ (radius of the circle) $O D=\frac{1}{\sqrt{2}} \quad$ (perpendicular distance) Using the sine function: $\sin (\theta)=\frac{O D}{O B}=\frac{\frac{1}{\sqrt{2}}}{1}=\frac{1}{\sqrt{2}}$ Thus, $\theta=45^{\circ}$.
Step 6: Find the Angle Subtended at the Center: The angle subtended at the center $A O B$ is twice the angle $D O B$ : $\angle A O B=2 \theta=2 \times 45^{\circ}=90^{\circ}$ In radians, this is: $\angle A O B=\frac{\pi}{2}$
Final Answer: The angle subtended by the chord $x+y=1$ at the center of the circle $x^2+y^2=1$ is $90^{\circ}$ or $\frac{\pi}{2}$ radians.
Asked in: AP EAMCET 2024 (19 May Shift 2)