If A, B are the points of contact of the tang | vedclass

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

Question

(B) The equation of the circle is $x^2+y^2-8x-10y+5=0$.Completing the square,we get $(x-4)^2+(y-5)^2 = 16+25-5 = 36 = 6^2$.Thus,the center $C$ is $(4,

Vedclass · Accepted Answer

$\frac{24}{7}$

Vedclass · Answer

(B) The equation of the circle is $x^2+y^2-8x-10y+5=0$.Completing the square,we get $(x-4)^2+(y-5)^2 = 16+25-5 = 36 = 6^2$.Thus,the center $C$ is $(4, 5)$ and the radius $r = 6$.The length of the tangent $PA$ from $P(-2, -3)$ is $\sqrt{S_1} = \sqrt{(-2)^2+(-3)^2-8(-2)-10(-3)+5} = \sqrt{4+9+16+30+5} = \sqrt{64} = 8$.In the right-angled triangle $	riangle CAP$,$\angle CAP = 90^\circ$.Let $\angle APC = \frac{	heta}{2}$. Then $	an(\frac{	heta}{2}) = \frac{CA}{PA} = \frac{6}{8} = \frac{3}{4}$.Using the formula $	an 	heta = \frac{2 	an(\frac{	heta}{2})}{1-	an^2(\frac{	heta}{2})}$,we have:$	an 	heta = \frac{2(\frac{3}{4})}{1-(\frac{3}{4})^2} = \frac{\frac{3}{2}}{1-\frac{9}{16}} = \frac{\frac{3}{2}}{\frac{7}{16}} = \frac{3}{2} 	imes \frac{16}{7} = \frac{24}{7}$.

If $A, B$ are the points of contact of the tangents drawn from the point $P(-2, -3)$ to the circle $x^2+y^2-8x-10y+5=0$ and the chord $AB$ subtends an angle $\theta$ at $P$,then $\tan \theta =$

Similar Questions

Let the tangents at two points $A$ and $B$ on the circle $x^{2} + y^{2} - 4x + 3 = 0$ meet at the origin $O(0,0)$. Then the area of the triangle $OAB$ is:

If the chord of contact of tangents from a point $A$ to a given circle passes through $B$,then the circle with $AB$ as a diameter will . . . . . .

The distance from the centre of the circle $x^2 + y^2 = 2x$ to the straight line passing through the points of intersection of the two circles $x^2 + y^2 + 5x - 8y + 1 = 0$ and $x^2 + y^2 - 3x + 7y - 25 = 0$ is-

If the chord of contact of the point $P(h, k)$ with respect to the circle $x^2+y^2-4x-4y+8=0$ meets the circle in two distinct points and it also makes an angle $45^{\circ}$ with the positive $X$-axis in the positive direction,then $(h, k)$ cannot be

Two circles $S_1 = px^2 + py^2 + 2g'x + 2f'y + d = 0$ and $S_2 = x^2 + y^2 + 2gx + 2fy + d' = 0$ have a common chord $PQ$. The equation of $PQ$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module