If a variable line,3x + 4y - lambda = 0,is su | vedclass

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

Question

(C) The centers of the circles are $C_1(1, 1)$ and $C_2(9, 1)$.Their radii are $r_1 = \sqrt{1^2 + 1^2 - 1} = 1$ and $r_2 = \sqrt{9^2 + 1^2 - 78} = \sq

Vedclass · Accepted Answer

$[12, 21]$

Vedclass · Answer

(C) The centers of the circles are $C_1(1, 1)$ and $C_2(9, 1)$.Their radii are $r_1 = \sqrt{1^2 + 1^2 - 1} = 1$ and $r_2 = \sqrt{9^2 + 1^2 - 78} = \sqrt{81 + 1 - 78} = 2$.For the circles to lie on opposite sides of the line $L: 3x + 4y - \lambda = 0$,the values of the expression $f(x, y) = 3x + 4y - \lambda$ at the centers must have opposite signs,and the line must not intersect the circles.$f(C_1) = 3(1) + 4(1) - \lambda = 7 - \lambda$.$f(C_2) = 3(9) + 4(1) - \lambda = 31 - \lambda$.Condition for opposite sides: $(7 - \lambda)(31 - \lambda) Also,the distance from the center to the line must be at least the radius:$d_1 = \frac{|3(1) + 4(1) - \lambda|}{\sqrt{3^2 + 4^2}} = \frac{|7 - \lambda|}{5} \ge 1 \implies |7 - \lambda| \ge 5 \implies \lambda \le 2$ or $\lambda \ge 12$.$d_2 = \frac{|3(9) + 4(1) - \lambda|}{\sqrt{3^2 + 4^2}} = \frac{|31 - \lambda|}{5} \ge 2 \implies |31 - \lambda| \ge 10 \implies \lambda \le 21$ or $\lambda \ge 41$.Taking the intersection of $\lambda \in (7, 31)$,$\lambda \in (-\infty, 2] \cup [12, \infty)$,and $\lambda \in (-\infty, 21] \cup [41, \infty)$,we get $\lambda \in [12, 21]$.

If a variable line,$3x + 4y - \lambda = 0$,is such that the two circles $x^2 + y^2 - 2x - 2y + 1 = 0$ and $x^2 + y^2 - 18x - 2y + 78 = 0$ are on its opposite sides,then the set of all values of $\lambda$ is the interval

Similar Questions

If the shortest distance between the parabola $y^2=4x$ and the circle $x^2+y^2-4x-16y+64=0$ is $d$,then $d^2$ is equal to:

The locus of the centre of the circle which touches the circle ${x^2} + {(y - 1)^2} = 1$ externally and also touches the $x$-axis is

Let $AB$ be a chord of the circle $x^2 + y^2 = r^2$ subtending a right angle at the centre. Then the locus of the centroid of the $\Delta PAB$ as $P$ moves on the circle is

Let $C$ be the largest circle centred at $(2,0)$ and inscribed in the ellipse $\frac{x^2}{36}+\frac{y^2}{16}=1$. If $(1, \alpha)$ lies on $C$,then $10 \alpha^2$ is equal to $.........$

The area (in $sq. units$) of the smaller of the two circles that touch the parabola $y^2 = 4x$ at the point $(1, 2)$ and the $x$-axis is

Vedclass Test Series

Exam Paper Generator

Online Exam Module