The length of the tangent drawn from the poin | vedclass

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

Question

(C) The equation of the circle is $2(x^2 + y^2) - 7x + 9y - 11 = 0$.Divide by $2$ to normalize the equation: $x^2 + y^2 - \frac{7}{2}x + \frac{9}{2}y

Vedclass · Accepted Answer

$\sqrt{14}$

Vedclass · Answer

(C) The equation of the circle is $2(x^2 + y^2) - 7x + 9y - 11 = 0$.Divide by $2$ to normalize the equation: $x^2 + y^2 - \frac{7}{2}x + \frac{9}{2}y - \frac{11}{2} = 0$.The length of the tangent from a point $(x_1, y_1)$ to a circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is given by $\sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c}$.Here,$x_1 = 2$,$y_1 = 3$,$g = -\frac{7}{4}$,$f = \frac{9}{4}$,and $c = -\frac{11}{2}$.Length of tangent $= \sqrt{2^2 + 3^2 - \frac{7}{2}(2) + \frac{9}{2}(3) - \frac{11}{2}}$.$= \sqrt{4 + 9 - 7 + \frac{27}{2} - \frac{11}{2}}$.$= \sqrt{6 + \frac{16}{2}} = \sqrt{6 + 8} = \sqrt{14}$.

The length of the tangent drawn from the point $(2, 3)$ to the circle $2(x^2 + y^2) - 7x + 9y - 11 = 0$ is:

Similar Questions

If a circle $C$ passing through the point $(4, 0)$ touches the circle $x^2+y^2+4x-6y=12$ externally at the point $(1, -1)$,then the radius of $C$ is

The circle $x^2 + y^2 + 4x - 4y + 4 = 0$ touches

The points $A(a, 0), B(0, b), C(c, 0)$ and $D(0, d)$ are such that $ac = bd$ and $a, b, c, d$ are all non-zero. Then the points:

The image of the point $(3, 4)$ with respect to the radical axis of the circles $x^2 + y^2 + 8x + 2y + 10 = 0$ and $x^2 + y^2 + 7x + 3y + 10 = 0$ is

The line $y = mx + c$ intersects the circle $x^2 + y^2 = r^2$ at two real distinct points,if

Vedclass Test Series

Exam Paper Generator

Online Exam Module