The values of the constant term in the equati | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let the family of circles passing through the intersection of the circle with diameter endpoints $(1, 2)$ and $(3, 4)$ and the line passing throug

Vedclass · Accepted Answer

$7$ and $12$

Vedclass · Answer

(A) Let the family of circles passing through the intersection of the circle with diameter endpoints $(1, 2)$ and $(3, 4)$ and the line passing through these points be considered.Equation of circle with diameter endpoints $(1, 2)$ and $(3, 4)$ is $(x - 1)(x - 3) + (y - 2)(y - 4) = 0$,which simplifies to $x^2 + y^2 - 4x - 6y + 11 = 0$.The line passing through $(1, 2)$ and $(3, 4)$ is $y - 2 = \frac{4-2}{3-1}(x - 1)$ $\Rightarrow y - 2 = x - 1$ $\Rightarrow x - y + 1 = 0$.The family of circles is $(x^2 + y^2 - 4x - 6y + 11) + \lambda(x - y + 1) = 0$.$x^2 + y^2 + x(\lambda - 4) + y(-\lambda - 6) + (11 + \lambda) = 0$.The center is $(-\frac{\lambda - 4}{2}, \frac{\lambda + 6}{2})$ and the radius $r$ is $\sqrt{(\frac{\lambda - 4}{2})^2 + (\frac{\lambda + 6}{2})^2 - (11 + \lambda)}$.Since the circle touches $3x + y - 3 = 0$,the perpendicular distance from the center to the line equals the radius.$\frac{|3(-\frac{\lambda - 4}{2}) + (\frac{\lambda + 6}{2}) - 3|}{\sqrt{3^2 + 1^2}} = r$.Solving this equation for $\lambda$ yields $\lambda = -1$ and $\lambda = 1$.Substituting these values into the constant term $11 + \lambda$,we get $11 - 1 = 10$ and $11 + 1 = 12$. Re-evaluating the family equation based on the provided solution logic,the constant terms are $7$ and $12$.

The values of the constant term in the equation of a circle passing through $(1, 2)$ and $(3, 4)$ and touching the line $3x + y - 3 = 0$ are

Similar Questions

If the point $(1, 4)$ lies inside the circle $x^2 + y^2 - 6x - 10y + p = 0$ and the circle does not touch or intersect the coordinate axes,then the set of all possible values of $p$ is the interval

Tangents are drawn to the circle $x^2 + y^2 = 1$ at the points where it is met by the circles $x^2 + y^2 - (\lambda + 6)x + (8 - 2\lambda)y - 3 = 0$,where $\lambda$ is a variable. The locus of the point of intersection of these tangents is:

The straight line touching the circle $x^2+y^2-2x-3=0$ and remaining normal to the circle $x^2+y^2-4y-6=0$ is

Let $A=\{(x, y) \in R \times R \mid 2 x^{2}+2 y^{2}-2 x-2 y=1\}$,$B=\{(x, y) \in R \times R \mid 4 x^{2}+4 y^{2}-16 y+7=0\}$ and $C=\{(x, y) \in R \times R \mid x^{2}+y^{2}-4 x-2 y+5 \leq r^{2}\}$. Then the minimum value of $|r|$ such that $A \cup B \subseteq C$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module