If a direct common tangent drawn to the circl | vedclass

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

Question

(D) The given equations of the circles are:$x^2 + y^2 - 6x + 4y + 9 = 0 \Rightarrow (x-3)^2 + (y+2)^2 = 2^2 \quad \dots(i)$$x^2 + y^2 + 2x - 2y + 1 =

Vedclass · Accepted Answer

$2\sqrt{6}$

Vedclass · Answer

(D) The given equations of the circles are:$x^2 + y^2 - 6x + 4y + 9 = 0 \Rightarrow (x-3)^2 + (y+2)^2 = 2^2 \quad \dots(i)$$x^2 + y^2 + 2x - 2y + 1 = 0 \Rightarrow (x+1)^2 + (y-1)^2 = 1^2 \quad \dots(ii)$From $(i)$ and $(ii)$,the centers and radii are:$C_1 = (3, -2), r_1 = 2$$C_2 = (-1, 1), r_2 = 1$Calculate the distance between the centers $C_1$ and $C_2$:$C_1C_2 = \sqrt{(3 - (-1))^2 + (-2 - 1)^2} = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = 5$The length of the direct common tangent $AB$ is given by the formula:$AB = \sqrt{(C_1C_2)^2 - (r_1 - r_2)^2}$$AB = \sqrt{5^2 - (2 - 1)^2} = \sqrt{25 - 1} = \sqrt{24} = 2\sqrt{6}$

If a direct common tangent drawn to the circles $x^2+y^2-6x+4y+9=0$ and $x^2+y^2+2x-2y+1=0$ touches the circles at $A$ and $B$,then $AB=$

Similar Questions

The diameter of a circle is $AB$ and $C$ is another point on the circle. Then the area of triangle $ABC$ will be:

If the inverse point of the point $(3, 2)$ with respect to the circle $x^2+y^2-2x+4y-4=0$ is $(l, m)$,then $(2l+19m) =$

If the straight line $y = mx$ is outside the circle $x^2 + y^2 - 20y + 90 = 0$,then

If the angle between the circles $x^2+y^2+4x-5=0$ and $x^2+y^2+2\lambda y-4=0$ is $\frac{\pi}{3}$,then $\lambda=$

Position of the point $(1, 1)$ with respect to the circle $x^2 + y^2 - x + y - 1 = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module