If the circle x^2 + y^2 = a^2 cuts a chord of | vedclass

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

Question

(B) The perpendicular distance $d$ from the center $(0,0)$ to the line $mx - y + c = 0$ is given by:$d = \frac{|m(0) - (0) + c|}{\sqrt{m^2 + (-1)^2}}

Vedclass · Accepted Answer

$(1 + m^2)(a^2 - b^2) = c^2$

Vedclass · Answer

(B) The perpendicular distance $d$ from the center $(0,0)$ to the line $mx - y + c = 0$ is given by:$d = \frac{|m(0) - (0) + c|}{\sqrt{m^2 + (-1)^2}} = \frac{|c|}{\sqrt{1 + m^2}}$In the right-angled triangle formed by the radius $a$,the half-chord $b$,and the perpendicular distance $d$,we have by Pythagoras theorem:$a^2 = b^2 + d^2$$d^2 = a^2 - b^2$Substituting the value of $d$:$\left(\frac{|c|}{\sqrt{1 + m^2}}\right)^2 = a^2 - b^2$$\frac{c^2}{1 + m^2} = a^2 - b^2$$c^2 = (1 + m^2)(a^2 - b^2)$

If the circle $x^2 + y^2 = a^2$ cuts a chord of length $2b$ on the line $y = mx + c$,then:

Similar Questions

The area of a circle whose centre is $(h, k)$ and radius $a$ is

If $5x - 12y + 10 = 0$ and $12y - 5x + 16 = 0$ are two tangents to a circle,then the radius of the circle is

The number of common tangents to the circles $x^2 + y^2 - 2x + 4y - 4 = 0$ and $x^2 + y^2 - 8x - 4y + 16 = 0$ is:

The length of the chord of the circle $x^{2}+y^{2}+3x+2y-8=0$ intercepted by the $y$-axis is

The equation of the chord of the circle $x^2+y^2-4x-10y+25=0$ having its midpoint at $(1,2)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module