From a point P on the line 4x - 3y = 6,two ta | vedclass

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

Question

(A) The given circle is $x^2 + y^2 - 6x - 4y + 4 = 0$. Its center is $C(3, 2)$ and radius $r = \sqrt{3^2 + 2^2 - 4} = \sqrt{9 + 4 - 4} = 3$. Let the a

Vedclass · Accepted Answer

$(6, 6)$

Vedclass · Answer

(A) The given circle is $x^2 + y^2 - 6x - 4y + 4 = 0$. Its center is $C(3, 2)$ and radius $r = \sqrt{3^2 + 2^2 - 4} = \sqrt{9 + 4 - 4} = 3$. Let the angle between the tangents be $2\alpha$. Then $2\alpha = 	an^{-1}\left(\frac{24}{7}ight)$,so $	an(2\alpha) = \frac{24}{7}$. Using $	an(2\alpha) = \frac{2	an\alpha}{1 - 	an^2\alpha} = \frac{24}{7}$,we get $12	an^2\alpha + 7	an\alpha - 12 = 0$. Solving this,we find $	an\alpha = 3/4$. In the right-angled triangle $	riangle PAC$,$\sin\alpha = \frac{r}{CP} = \frac{3}{CP}$. Since $	an\alpha = 3/4$,we have $\sin\alpha = 3/5$. Thus,$\frac{3}{CP} = \frac{3}{5} \Rightarrow CP = 5$. If $P = (h, k)$,then $(h-3)^2 + (k-2)^2 = 5^2 = 25$,which simplifies to $h^2 + k^2 - 6h - 4k - 12 = 0$. Since $P$ lies on $4x - 3y = 6$,$h = \frac{6 + 3k}{4}$. Substituting this into the circle equation gives $k^2 - 4k - 12 = 0$,so $(k-6)(k+2) = 0$. If $k = 6$,$h = 6$. If $k = -2$,$h = 0$. Thus,$P$ can be $(6, 6)$ or $(0, -2)$.

From a point $P$ on the line $4x - 3y = 6$,two tangents are drawn to the circle $x^2 + y^2 - 6x - 4y + 4 = 0$. If the angle between these tangents is $\tan^{-1}\left(\frac{24}{7}\right)$,then $P$ can be:

Similar Questions

The equation of the circle circumscribing the triangle formed by the straight lines $x+y=6$,$2x+y=4$,and $x+2y=5$ is given by:

$A$ circle $C_1$ of radius $2$ touches both $x$-axis and $y$-axis. Another circle $C_2$ whose radius is greater than $2$ touches circle $C_1$ and both the axes. Then the radius of circle $C_2$ is

If the midpoint of the chord intercepted by the circle $x^2+y^2-8x+10y+5=0$ on the line $2x+y+2=0$ is $(h, k)$,then $k+4h=$

Consider a circle $C_1: x^2+y^2-4x-2y=\alpha-5$. Let its mirror image in the line $y=2x+1$ be another circle $C_2: 5x^2+5y^2-10fx-10gy+36=0$. Let $r$ be the radius of $C_2$. Then $\alpha+r$ is equal to $......$.

$A$ cow is tied to a post by a rope. The cow moves along a circular path,always keeping the rope tight. If it describes an arc of $44 \ m$ when it has traced out $72^{\circ}$ at the centre,the length of the rope is: (in $m$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module