If OA and OB are tangents from the origin to | vedclass

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

Question

(B) The equation of the circle is $x^{2} + y^{2} - 6x - 8y + 21 = 0$. The center is $C(3, 4)$ and the radius $r = \sqrt{3^{2} + 4^{2} - 21} = \sqrt{9

Vedclass · Accepted Answer

$\frac{4}{5}\sqrt{21}$

Vedclass · Answer

(B) The equation of the circle is $x^{2} + y^{2} - 6x - 8y + 21 = 0$. The center is $C(3, 4)$ and the radius $r = \sqrt{3^{2} + 4^{2} - 21} = \sqrt{9 + 16 - 21} = \sqrt{4} = 2$.The equation of the chord of contact $AB$ from the origin $(0, 0)$ is given by $T = 0$,which is $x(0) + y(0) - 3(x + 0) - 4(y + 0) + 21 = 0$,simplifying to $3x + 4y - 21 = 0$.The perpendicular distance $CM$ from the center $C(3, 4)$ to the line $3x + 4y - 21 = 0$ is $CM = \frac{|3(3) + 4(4) - 21|}{\sqrt{3^{2} + 4^{2}}} = \frac{|9 + 16 - 21|}{5} = \frac{4}{5}$.In the right-angled triangle $	riangle AMC$,$AM = \sqrt{AC^{2} - CM^{2}} = \sqrt{2^{2} - (\frac{4}{5})^{2}} = \sqrt{4 - \frac{16}{25}} = \sqrt{\frac{100 - 16}{25}} = \sqrt{\frac{84}{25}} = \frac{2\sqrt{21}}{5}$.Since $AB = 2AM$,we have $AB = 2 	imes \frac{2\sqrt{21}}{5} = \frac{4}{5}\sqrt{21}$.

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

Similar Questions

If the midpoint of a chord of the circle $x^2 + y^2 + x - y - 1 = 0$ is $(1, 1)$,then the length of the chord is

Tangents are drawn from the point $(4, 3)$ to the circle $x^2 + y^2 = 9$. The area of the triangle formed by them and the line joining their points of contact is

The length of the common chord of the circles $x^2 + y^2 - 6x - 16 = 0$ and $x^2 + y^2 - 8y - 9 = 0$ is:

The point of intersection of the tangents drawn at the points where the line $2x - y + 3 = 0$ meets the circle $x^2 + y^2 - 4x - 6y + 4 = 0$ is

If the chord of contact of the point $P(1, 1)$ with respect to the circle $S = x^2 + y^2 + 4x + 6y - 3 = 0$ meets the circle $S = 0$ at $A$ and $B$,then the area of $\triangle PAB$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module