Question:
If the point \((2,\lambda)\) lies inside the circles \(x^2+y^2=13\) and \(x^2+y^2+x-2y=14\), then \(\lambda\) lies in the set:
If the point \((2,\lambda)\) lies inside the circles \(x^2+y^2=13\) and \(x^2+y^2+x-2y=14\), then \(\lambda\) lies in the set:
Show Hint
To check whether a point lies inside a circle, substitute the point in the circle equation and use the inequality form. For a circle \(S=0\), the inside region is generally obtained by comparing with the radius form or using the corresponding inequality.
Updated On: Jun 26, 2026
- \((-\infty,-3)\cup(4,\infty)\)
- \((-\infty,-1)\cup(3,\infty)\)
- \([-3,4]\)
- \([-2,3]\)
Show Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Step 1: Use the condition for a point to lie inside the first circle.
The first circle is \[ x^2+y^2=13 \] For the point \((2,\lambda)\), we substitute \[ x=2,\quad y=\lambda \] For the point to lie inside or on the circle, \[ 2^2+\lambda^2\leq 13 \] \[ 4+\lambda^2\leq 13 \] \[ \lambda^2\leq 9 \] Hence, \[ -3\leq \lambda \leq 3 \]
Step 2: Use the condition for a point to lie inside the second circle.
The second circle is \[ x^2+y^2+x-2y=14 \] Substitute \[ x=2,\quad y=\lambda \] So, \[ 2^2+\lambda^2+2-2\lambda\leq 14 \] \[ 4+\lambda^2+2-2\lambda\leq 14 \] \[ \lambda^2-2\lambda+6\leq 14 \] \[ \lambda^2-2\lambda-8\leq 0 \] Factorizing, \[ (\lambda-4)(\lambda+2)\leq 0 \] Thus, \[ -2\leq \lambda \leq 4 \]
Step 3: Take the common interval.
From the first circle, \[ \lambda\in[-3,3] \] From the second circle, \[ \lambda\in[-2,4] \] Therefore, the common interval is \[ [-3,3]\cap[-2,4]=[-2,3] \]
Step 4: Final conclusion.
Hence, \[ \boxed{[-2,3]} \]
The first circle is \[ x^2+y^2=13 \] For the point \((2,\lambda)\), we substitute \[ x=2,\quad y=\lambda \] For the point to lie inside or on the circle, \[ 2^2+\lambda^2\leq 13 \] \[ 4+\lambda^2\leq 13 \] \[ \lambda^2\leq 9 \] Hence, \[ -3\leq \lambda \leq 3 \]
Step 2: Use the condition for a point to lie inside the second circle.
The second circle is \[ x^2+y^2+x-2y=14 \] Substitute \[ x=2,\quad y=\lambda \] So, \[ 2^2+\lambda^2+2-2\lambda\leq 14 \] \[ 4+\lambda^2+2-2\lambda\leq 14 \] \[ \lambda^2-2\lambda+6\leq 14 \] \[ \lambda^2-2\lambda-8\leq 0 \] Factorizing, \[ (\lambda-4)(\lambda+2)\leq 0 \] Thus, \[ -2\leq \lambda \leq 4 \]
Step 3: Take the common interval.
From the first circle, \[ \lambda\in[-3,3] \] From the second circle, \[ \lambda\in[-2,4] \] Therefore, the common interval is \[ [-3,3]\cap[-2,4]=[-2,3] \]
Step 4: Final conclusion.
Hence, \[ \boxed{[-2,3]} \]
Was this answer helpful?
0
0
Top AP EAPCET Mathematics Questions
- $D = \left\{x\in\mathbb{R}: f\left(x\right) =\sqrt{\frac{x - \left|x\right|}{x - \left[x\right]}} \text{is defined} \right\}$ and $C$ be the range of the real function $g(x) = \frac{2x}{4 + x^{2}}$. Then $D \cap C$
- $f\left(x\right) = \frac{x}{e^{x}-1} + \frac{x}{2} + 2 \cos^{3} \frac{x}{2} $ on $R-\left\{0\right\} $ is
- Consider the following lists.
List-I List-II (A) $f(x) = \frac{|x+2|}{x+2} , x \ne -2 $ (I) $[\frac{1}{3} , 1 ]$ (B) $(x)=|[x]|,x \in [R$ (II) Z (C) $h(x) = |x - [x]| , x \in [R$ (III) W (D) $f(x) = \frac{1}{2 - \sin 3x} , x \in [R$ (IV) [0, 1) (V) { -1, 1} - If $[x]$ is the greatest integer less than or equal to $x$ and $|x|$ is the modulus of $x$. then the system of three equations $2x + 3 | y | + 5[z] = 0, x + |y| - 2[z] = 4, x + |y| + |z| = 1$ has
- Investigate the values of $\lambda$ and $\mu$ for the system $x+2 y+3 z=6, x+3 y+5 z=9$, $2 x+5 y+\lambda z=\mu$ and match the values in List - I with the items in List - II.
List I List II (A) $\lambda=8, \mu \neq 15$ 1. Infinitely many solutions (B) $\lambda \neq 8, \mu \in R$ 2. No solution (C) $\lambda=8, \mu=15$ 3. Unique solution
View More Questions
Top AP EAPCET Circles Questions
- If the equation of the circle whose radius is 3 units and which touches internally the circle $$ x^2 + y^2 - 4x - 6y - 12 = 0 $$ at the point $ (-1, -1) $ is $$ x^2 + y^2 + px + qy + r = 0, $$ then $ p + q - r $ is:
- The equation of the circle touching the circle \[ x^2 + y^2 - 6x + 6y + 17 = 0 \] externally and to which the lines \[ x^2 - 3xy - 3x + 9y = 0 \] are normal is:
The equation of a circle which touches the straight lines $x + y = 2$, $x - y = 2$ and also touches the circle $x^2 + y^2 = 1$ is:
- The radical axis of the circles \[ x^2 + y^2 + 2gx + 2fy + c = 0 \] and \[ 2x^2 + 2y^2 + 3x + 8y + 2c = 0 \] touches the circle \[ x^2 + y^2 + 2x + 2y + 1 = 0. \] Then:
- A circle touches the y-axis at a distance 4 units from the origin and cuts off an intercept 6 from the x-axis. Find its equation.
View More Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions