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

Let $x_0, y_0$ be fixed real numbers such that $x_0^2+y_0^2 > 1$. If $x, y$ are arbitrary real numbers such that $x^2+y^2 \leq 1$,then the minimum value of $(x-x_0)^2+(y-y_0)^2$ is

If a circle passes through the points $(0,0), (x,0),$ and $(0,y)$,then the coordinates of its centre are

If the equation of the circle whose radius is $3$ units and which touches internally the circle $x^2+y^2-4x-6y-12=0$ at the point $(-1,-1)$ is $x^2+y^2+px+qy+r=0$,then $p+q-r=$

The equation of the circle which touches the coordinate axes and the line $\frac{x}{3} + \frac{y}{4} = 1$,and whose centre lies in the first quadrant,is ${x^2} + {y^2} - 2cx - 2cy + {c^2} = 0$,where $c$ is:

$A$ line $l$ meets the circle $x^2+y^2=61$ at points $A$ and $B$. Given that $P(-5, 6)$ is a point such that $PA=PB=10$,find the equation of line $l$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module