The perpendicular distance from the point (1, | vedclass

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

Question

(B) Given circles are $S_1: x^2+y^2-2x+4y-4=0$ and $S_2: x^2+y^2+4x-6y-3=0$. The equation of the common chord is given by $S_1 - S_2 = 0$. $(x^2+y^2-2

Vedclass · Accepted Answer

$\frac{13}{\sqrt{136}}$

Vedclass · Answer

(B) Given circles are $S_1: x^2+y^2-2x+4y-4=0$ and $S_2: x^2+y^2+4x-6y-3=0$. The equation of the common chord is given by $S_1 - S_2 = 0$. $(x^2+y^2-2x+4y-4) - (x^2+y^2+4x-6y-3) = 0$. $-6x + 10y - 1 = 0$,which simplifies to $6x - 10y + 1 = 0$. The perpendicular distance from the point $(x_1, y_1) = (1, 2)$ to the line $Ax + By + C = 0$ is given by $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$. Substituting the values: $d = \frac{|6(1) - 10(2) + 1|}{\sqrt{6^2 + (-10)^2}}$. $d = \frac{|6 - 20 + 1|}{\sqrt{36 + 100}} = \frac{|-13|}{\sqrt{136}} = \frac{13}{\sqrt{136}}$ units.

The perpendicular distance from the point $(1,2)$ to the common chord of the circles $x^2+y^2-2x+4y-4=0$ and $x^2+y^2+4x-6y-3=0$ is ........ units.

Similar Questions

If the common chord of the circles $x^2+y^2+4y=0$ and $x^2+y^2-4x-5=0$ is the diameter of the circle $S=0$,then the abscissa of the centre of the circle $S=0$ is

If $A$ and $B$ are the points of intersection of the circles $x^2+y^2-4x+6y-3=0$ and $x^2+y^2+2x-2y-2=0$,then the distance between $A$ and $B$ is

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:

If $OA$ and $OB$ are tangents from the origin to the circle $x^{2} + y^{2} - 6x - 8y + 21 = 0$,then $AB = \dots$

The intercept on the line $y = x$ by the circle ${x^2} + {y^2} - 2x = 0$ is $AB$. The equation of the circle having $AB$ as a diameter is

Vedclass Test Series

Exam Paper Generator

Online Exam Module