Consider L_1: 2x + 3y + p - 3 = 0; L_2: 2x + | vedclass

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

Question

(C) The equation of the circle $C$ is $x^2 + 6x + 9 + y^2 - 10y + 25 = -30 + 9 + 25$,which simplifies to $(x + 3)^2 + (y - 5)^2 = 4$.The center of the

Vedclass · Accepted Answer

$STATEMENT-1$ is True,$STATEMENT-2$ is False.

Vedclass · Answer

(C) The equation of the circle $C$ is $x^2 + 6x + 9 + y^2 - 10y + 25 = -30 + 9 + 25$,which simplifies to $(x + 3)^2 + (y - 5)^2 = 4$.The center of the circle is $(-3, 5)$ and the radius $r = 2$.For $L_1: 2x + 3y + p - 3 = 0$ to be a chord,the distance from the center $(-3, 5)$ to $L_1$ must be less than the radius $r=2$.Distance $d_1 = \frac{|2(-3) + 3(5) + p - 3|}{\sqrt{2^2 + 3^2}} = \frac{|p + 6|}{\sqrt{13}} For $L_2: 2x + 3y + p + 3 = 0$ to be a diameter,it must pass through the center $(-3, 5)$,so $2(-3) + 3(5) + p + 3 = 0 \Rightarrow p = -12$.If $p = -12$,$L_1$ becomes $2x + 3y - 15 = 0$. The distance $d_1 = \frac{|-12 + 6|}{\sqrt{13}} = \frac{6}{\sqrt{13}} \approx 1.66 For $STATEMENT-2$: If $L_1$ is a diameter,$2(-3) + 3(5) + p - 3 = 0 \Rightarrow p = -6$. Then $L_2$ is $2x + 3y - 3 = 0$. The distance from the center to $L_2$ is $d_2 = \frac{|2(-3) + 3(5) - 3|}{\sqrt{13}} = \frac{6}{\sqrt{13}} \approx 1.66 < 2$. Since $d_2 < r$,$L_2$ is a chord. Thus,$STATEMENT-2$ is False.

Similar Questions

If the circles $x^2+y^2-2x-2y+k=0$ and $x^2+y^2+4x+6y+4=0$ touch each other externally,then the point of contact of the two circles is

Suppose $d_1$ and $d_2$ are respectively the lengths of intercepts of the circles $x^2+y^2=4$ and $x^2+y^2-10x-14y+65=0$ on the line $2x-2y-3=0$. Then,which of the following is true?

The equation of the circle which touches the circle $S \equiv x^2+y^2-10x-4y+19=0$ at the point $(2,3)$ internally and has a radius equal to half of the radius of the circle $S=0$ is:

Suppose the tangents drawn to the circle $x^2+y^2-6x-4y-11=0$ from $P(1,8)$ touch the circle at $A$ and $B$. Then the centre of the circle passing through $P, A$ and $B$ is

Among all the parallelograms whose diagonals are $10$ and $4$,the one having maximum area has its perimeter lying in the interval

Vedclass Test Series

Exam Paper Generator

Online Exam Module