If the tangents at the points P and Q on the | vedclass

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

Question

(D) The equation of the circle is $x^2 + y^2 - 2x + y - 5 = 0$.Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -1$,$f = \frac{1}{2}$,and $c

Vedclass · Accepted Answer

$\frac{5}{8}$

Vedclass · Answer

(D) The equation of the circle is $x^2 + y^2 - 2x + y - 5 = 0$.Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -1$,$f = \frac{1}{2}$,and $c = -5$.The center of the circle is $C(-g, -f) = (1, -\frac{1}{2})$.The radius $r$ is $\sqrt{g^2 + f^2 - c} = \sqrt{1 + \frac{1}{4} + 5} = \sqrt{\frac{25}{4}} = \frac{5}{2}$.Let $L$ be the length of the tangent $PR = QR$. $L = \sqrt{S_1} = \sqrt{(\frac{9}{4})^2 + 2^2 - 2(\frac{9}{4}) + 2 - 5} = \sqrt{\frac{81}{16} + 4 - \frac{9}{2} + 2 - 5} = \sqrt{\frac{81 - 72 + 16}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}$.The distance $d$ from the center $C$ to the point $R$ is $d = \sqrt{(\frac{9}{4} - 1)^2 + (2 + \frac{1}{2})^2} = \sqrt{(\frac{5}{4})^2 + (\frac{5}{2})^2} = \sqrt{\frac{25}{16} + \frac{25}{4}} = \sqrt{\frac{125}{16}} = \frac{5\sqrt{5}}{4}$.The area of $	riangle PQR$ is given by $\frac{r L^3}{r^2 + L^2} = \frac{(\frac{5}{2}) \cdot (\frac{5}{4})^3}{(\frac{5}{2})^2 + (\frac{5}{4})^2} = \frac{\frac{5}{2} \cdot \frac{125}{64}}{\frac{25}{4} + \frac{25}{16}} = \frac{\frac{625}{128}}{\frac{100 + 25}{16}} = \frac{625}{128} \cdot \frac{16}{125} = \frac{5}{8}$.

If the tangents at the points $P$ and $Q$ on the circle $x^2 + y^2 - 2x + y = 5$ meet at the point $R \left(\frac{9}{4}, 2\right)$,then the area of the triangle $PQR$ is

Similar Questions

If $OA$ and $OB$ are the tangents from the origin to the circle ${x^2} + {y^2} + 2gx + 2fy + c = 0$ and $C$ is the centre of the circle,the area of the quadrilateral $OACB$ is

The length of the tangent from the point $(4, 5)$ to the circle $x^2 + y^2 + 2x - 6y - 6 = 0$ is

Two tangents to the circle $x^2+y^2=4$ at the points $A$ and $B$ meet at $P(-4,0)$. Then the area of quadrilateral $PAOB$,where $O$ is the origin,is

For what possible value of $p$ is the line $x \cos \alpha + y \sin \alpha = p$ a tangent to the circle $x^2 + y^2 - 2qx \cos \alpha - 2qy \sin \alpha = 0$?

The point on the curve ${y^2} = 2(x - 3)$ at which the normal is parallel to the line $y - 2x + 1 = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module