The length of the common chord of the two cir | vedclass

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

Question

(A) The equations of the circles are:$S_1: x^2 + y^2 + \frac{7}{2}x - \frac{5}{2}y + 1 = 0$ (Dividing by $2$)$S_2: x^2 + y^2 - 4x + 8y - 18 = 0$The eq

Vedclass · Accepted Answer

$2 \sqrt{\frac{1102}{333}}$

Vedclass · Answer

(A) The equations of the circles are:$S_1: x^2 + y^2 + \frac{7}{2}x - \frac{5}{2}y + 1 = 0$ (Dividing by $2$)$S_2: x^2 + y^2 - 4x + 8y - 18 = 0$The equation of the common chord is $S_1 - S_2 = 0$:$(x^2 + y^2 + \frac{7}{2}x - \frac{5}{2}y + 1) - (x^2 + y^2 - 4x + 8y - 18) = 0$$\frac{15}{2}x - \frac{21}{2}y + 19 = 0 \Rightarrow 15x - 21y + 38 = 0$The center of circle $S_2$ is $C_2(2, -4)$ and its radius $r = \sqrt{2^2 + (-4)^2 - (-18)} = \sqrt{4 + 16 + 18} = \sqrt{38}$.The perpendicular distance $d$ from $C_2(2, -4)$ to the line $15x - 21y + 38 = 0$ is:$d = \frac{|15(2) - 21(-4) + 38|}{\sqrt{15^2 + (-21)^2}} = \frac{|30 + 84 + 38|}{\sqrt{225 + 441}} = \frac{152}{\sqrt{666}}$.The length of the common chord is $2\sqrt{r^2 - d^2}$:$2\sqrt{38 - \frac{152^2}{666}} = 2\sqrt{38 - \frac{23104}{666}} = 2\sqrt{\frac{25308 - 23104}{666}} = 2\sqrt{\frac{2204}{666}} = 2\sqrt{\frac{1102}{333}}$.

The length of the common chord of the two circles $2x^{2} + 2y^{2} + 7x - 5y + 2 = 0$ and $x^{2} + y^{2} - 4x + 8y - 18 = 0$ is:

Similar Questions

The combined equation of the direct common tangents of the circles $x^2+y^2-2x-2y-2=0$ and $x^2+y^2+4x+6y+12=0$ is:

If the circle $C_1 : x^{2} + y^{2} = 16$ intersects another circle $C_2$ of radius $5$ such that the length of the common chord is maximum and its slope is $3/4$,then the coordinates of the center of $C_2$ are:

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

If the angle of intersection at a point where the two circles with radii $5 \ cm$ and $12 \ cm$ intersect is $90^o$,then the length (in $cm$) of their common chord is

If $P$ and $Q$ are the points of intersection of the circles $x^2 + y^2 + 3x + 7y + 2p - 5 = 0$ and $x^2 + y^2 + 2x + 2y - p^2 = 0$,then there is a circle passing through $P, Q$ and $(1, 1)$ for:

Vedclass Test Series

Exam Paper Generator

Online Exam Module