Let 'C' be the locus of the centre of circles | vedclass

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

Question

(B) The locus of the centre of circles touching two lines is the pair of angle bisectors of the given lines.Given lines are $L_1: x + 2y - 5 = 0$ and

Vedclass · Accepted Answer

$59$

Vedclass · Answer

(B) The locus of the centre of circles touching two lines is the pair of angle bisectors of the given lines.Given lines are $L_1: x + 2y - 5 = 0$ and $L_2: 2x - 4y + 7 = 0$.The angle bisectors are given by $\frac{x + 2y - 5}{\sqrt{1^2 + 2^2}} = \pm \frac{2x - 4y + 7}{\sqrt{2^2 + (-4)^2}}$.$\frac{x + 2y - 5}{\sqrt{5}} = \pm \frac{2x - 4y + 7}{2\sqrt{5}}$.$2(x + 2y - 5) = \pm (2x - 4y + 7)$.Case $1$: $2x + 4y - 10 = 2x - 4y + 7 \Rightarrow 8y = 17 \Rightarrow y = \frac{17}{8}$.Case $2$: $2x + 4y - 10 = -2x + 4y - 7 \Rightarrow 4x = 3 \Rightarrow x = \frac{3}{4}$.The locus $'C'$ consists of the lines $x = \frac{3}{4}$ and $y = \frac{17}{8}$.These lines intersect at $(\frac{3}{4}, \frac{17}{8})$.We need the area enclosed by these two lines and the line $x - y - 5 = 0$.The intersection points are:$1$) Intersection of $x = \frac{3}{4}$ and $x - y - 5 = 0$: $\frac{3}{4} - y - 5 = 0 \Rightarrow y = \frac{3}{4} - 5 = -\frac{17}{4}$. Point $A = (\frac{3}{4}, -\frac{17}{4})$.$2$) Intersection of $y = \frac{17}{8}$ and $x - y - 5 = 0$: $x - \frac{17}{8} - 5 = 0 \Rightarrow x = \frac{17}{8} + 5 = \frac{57}{8}$. Point $B = (\frac{57}{8}, \frac{17}{8})$.$3$) Intersection of $x = \frac{3}{4}$ and $y = \frac{17}{8}$: Point $V = (\frac{3}{4}, \frac{17}{8})$.The area of the right-angled triangle formed by these vertices is $\frac{1}{2} 	imes |x_B - x_V| 	imes |y_V - y_A|$.$|x_B - x_V| = |\frac{57}{8} - \frac{6}{8}| = \frac{51}{8}$.$|y_V - y_A| = |\frac{17}{8} - (-\frac{34}{8})| = \frac{51}{8}$.Area $= \frac{1}{2} 	imes \frac{51}{8} 	imes \frac{51}{8} = \frac{51^2}{2 	imes 8^2}$.Comparing with $\frac{P^2}{2Q^2}$,we get $P = 51$ and $Q = 8$.Since $P$ and $Q$ are relative primes,$P + Q = 51 + 8 = 59$.

Let $'C'$ be the locus of the centre of circles which touch the lines $x + 2y - 5 = 0$ and $2x - 4y + 7 = 0$. If the area enclosed by the curve $'C'$ with the line $x - y - 5 = 0$ is $\frac{P^2}{2Q^2}$,where $P$ and $Q$ are relative primes,then the value of $P + Q$ is:

Similar Questions

The equations of the angle bisectors between the lines $3x - 4y + 7 = 0$ and $12x - 5y - 8 = 0$ are:

The straight line $x+y+1=0$ bisects an angle between the pair of lines of which one is $2x+3y-4=0$. Then,the equation of the other line is

The family of lines,forming an isosceles triangle with the lines $3x - 4y - 2 = 0$ and $12x - 5y + 6 = 0$,is

Statement $-I$: Two lines which pass through a given fixed point and are equally inclined to two other lines passing through the same point,are always perpendicular to each other.
Statement $-II$: Angle bisectors of two intersecting lines are always perpendicular to each other.

Vedclass Test Series

Exam Paper Generator

Online Exam Module