The points of intersection of the line $a x+b y=0,(a \neq b)$ and the circle $x^2+y^2-2 x=0$ are $A (\alpha, 0)$ and $B (1, \beta)$ The image of the circle with $AB$ as a diameter in the line $x+y+2=0$ is :
- $x^2+y^2+3 x+3 y+4=0$
- $x^2+y^2+5 x+5 y+12=0$
- $x^2+y^2+3 x+5 y+8=0$
- $x^2+y^2-5 x-5 y+12=0$
The Correct Option is B
Solution and Explanation
\(\text{Only possibility}\) \(α=0,β=1 \)
\(∴\text{ equation of circle}\; x^2+y^2−x−y=0\)
\(\text{Image of circle in}\; x+y+2=0 \;is \)
\(x^2+y^2+5x+5y+12=0\)
\(\text{Hence, the correct option is(B):}\) \(x^2+y^2+5x+5y+12=0\)
Top Questions on Circle
Four distinct points \( (2k, 3k), (1, 0), (0, 1) \) and \( (0, 0) \) lie on a circle for \( k \) equal to:
- If the circles \( (x+1)^2 + (y+2)^2 = r^2 \) and \( x^2 + y^2 - 4x - 4y + 4 = 0 \) intersect at exactly two distinct points, then
- If one of the diameters of the circle \(x^2+y^2-10x+4y+13=0\) is a chord of another circle and whose center is the point of intersection of the lines \(2x + 3y = 12\) and \(3x - 2y = 5 \) then the radius of the circle is
- Let C be the circle in the complex plane with centre z0 =\(\frac{1}{2}\) (1+3i) and radius r = 1. Let z1 = 1+ i and the complex number z2 be outside the circle C such that |z1 – z0| |z2 – z0| = 1. If z0, z1 and z2 are collinear, then the smaller value of |z2|2 is equal to
- The locus of the mid points of the chords of the circle $C_1:(x-4)^2+(y-5)^2=4$ which subtend an angle $\theta_i$ at the centre of the circle $C_1$, is a circle of radius $r_i$ If $\theta_1=\frac{\pi}{3}, \theta_3=\frac{2 \pi}{3}$ and $r_1^2=r_2^2+r_3^2$, then $\theta_2$ is equal to
Questions Asked in JEE Main exam
- Let \[ \vec{a} = 2\hat{i} + \alpha \hat{j} + \hat{k}, \quad \vec{b} = -\hat{i} + \hat{k}, \quad \vec{c} = \beta \hat{j} - \hat{k}, \] where \( \alpha \) and \( \beta \) are integers and \( \alpha \beta = -6 \). Let the values of the ordered pair \( (\alpha, \beta) \) for which the area of the parallelogram of diagonals \( \vec{a} + \vec{b} \) and \( \vec{b} + \vec{c} \) is \( \frac{\sqrt{21}}{2} \), be \( (\alpha_1, \beta_1) \) and \( (\alpha_2, \beta_2) \). Then \( \alpha_1^2 + \beta_1^2 - \alpha_2 \beta_2 \) is equal to:
- JEE Main - 2024
- Vectors
- The molecular formula of second homologue in the homologous series of monocarboxylic acid is
- JEE Main - 2024
- Aldehydes, Ketones and Carboxylic Acids
- Given below are two statements: Statement I: $\text{PF}_5$ and $\text{BrF}_5$ both exhibit $\text{sp}^3\text{d}$ hybridisation.Statement II: Both $\text{SF}_6$ and $[\text{Co}(\text{NH}_3)_6]^{3+}$ exhibit $\text{sp}^3\text{d}^2$ hybridisation. In the light of the above statements,choose the correct answer from the options given below:
- JEE Main - 2024
- Hybridisation
- Consider the system of linear equations \( x + y + z = 4\mu \), \( x + 2y + 2\lambda z = 10\mu \), \( x + 3y + 4\lambda^2 z = \mu^2 + 15 \), where \( \lambda, \mu \in \mathbb{R} \). Which one of the following statements is NOT correct?
- JEE Main - 2024
- Linear Algebra
- If the constant term in the expansion of \((1 + 2x - 3x^3) \left(\frac{3}{2}x^2 - \frac{1}{3x}\right)^9\) is \( p \), then \( 108p \) is equal to _________.
- JEE Main - 2024
- Algebra
Concepts Used:
Circle
A circle can be geometrically defined as a combination of all the points which lie at an equal distance from a fixed point called the centre. The concepts of the circle are very important in building a strong foundation in units likes mensuration and coordinate geometry. We use circle formulas in order to calculate the area, diameter, and circumference of a circle. The length between any point on the circle and its centre is its radius.
Any line that passes through the centre of the circle and connects two points of the circle is the diameter of the circle. The radius is half the length of the diameter of the circle. The area of the circle describes the amount of space that is covered by the circle and the circumference is the length of the boundary of the circle.
Also Check: