The shortest distance from the line 3x + 4y = | vedclass

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

Question

(B) The equation of the circle is $x^2 + y^2 - 6x + 8y = 0$. Comparing this with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get the center $C = (-g, -f) = (3,

Vedclass · Accepted Answer

$\frac{7}{5}$

Vedclass · Answer

(B) The equation of the circle is $x^2 + y^2 - 6x + 8y = 0$. Comparing this with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get the center $C = (-g, -f) = (3, -4)$ and radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{3^2 + (-4)^2 - 0} = \sqrt{9 + 16} = 5$. The perpendicular distance $d$ from the center $(3, -4)$ to the line $3x + 4y - 25 = 0$ is given by $d = \frac{|3(3) + 4(-4) - 25|}{\sqrt{3^2 + 4^2}} = \frac{|9 - 16 - 25|}{\sqrt{25}} = \frac{|-32|}{5} = \frac{32}{5}$. The shortest distance from the line to the circle is $d - r = \frac{32}{5} - 5 = \frac{32 - 25}{5} = \frac{7}{5}$.

The shortest distance from the line $3x + 4y = 25$ to the circle $x^2 + y^2 - 6x + 8y = 0$ is

Similar Questions

The points $(x_1, y_1), (x_2, y_2), (x_1, y_2),$ and $(x_2, y_1)$ are always:

Let $l > 0$ be a real number,$C$ denote a circle with circumference $l$,and $T$ denote a triangle with perimeter $l$. Then:

$A$ circle touches both the coordinate axes and the straight line $L \equiv 4x+3y-6=0$ in the first quadrant. If this circle lies below the line $L=0$,then the equation of that circle is

For the circle $x^2 + y^2 - 2x + 4y - 4 = 0$,what is the line $2x - y - 1 = 0$?

If the point $(2 \cos \theta, 2 \sin \theta)$ for $\theta \in (0, 2 \pi)$ lies in the region between the lines $x+y=2$ and $x-y=2$ containing the origin,then $\theta$ lies in

Vedclass Test Series

Exam Paper Generator

Online Exam Module