Let O be the centre of the circle x^2 + y^2 = | vedclass

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

Question

(B) Let the centre of the circumcircle of $	riangle OPQ$ be $C(h, k)$.Since $O(0, 0)$ is a vertex of the triangle,the circumcircle passes through the

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Let the centre of the circumcircle of $	riangle OPQ$ be $C(h, k)$.Since $O(0, 0)$ is a vertex of the triangle,the circumcircle passes through the origin.The equation of the chord $PQ$ is $2x + 4y = 5$.The equation of the chord of the circle $x^2 + y^2 = r^2$ with midpoint $(h, k)$ is $hx + ky = h^2 + k^2$.Comparing $2x + 4y = 5$ with $hx + ky = r^2$ (since the chord $PQ$ is the polar of the centre $C$ with respect to the circle $x^2 + y^2 = r^2$ is not directly applicable,we use the property that the line $OC$ is perpendicular to $PQ$ and the midpoint of $PQ$ lies on $OC$):The slope of $PQ$ is $m = -\frac{2}{4} = -\frac{1}{2}$.Since $OC \perp PQ$,the slope of $OC$ is $2$.The line $OC$ passes through $(0, 0)$,so its equation is $y = 2x$.The centre $C(h, k)$ lies on $y = 2x$,so $k = 2h$.Also,$C(h, k)$ lies on the given line $x + 2y = 4$.Substituting $k = 2h$ into $x + 2y = 4$,we get $h + 2(2h) = 4 \Rightarrow 5h = 4 \Rightarrow h = \frac{4}{5}$.Thus,$k = 2(\frac{4}{5}) = \frac{8}{5}$,so $C = (\frac{4}{5}, \frac{8}{5})$.Since $C$ is the circumcentre of $	riangle OPQ$,$CO = CP = CQ = r_{circum}$.$CO^2 = (\frac{4}{5})^2 + (\frac{8}{5})^2 = \frac{16+64}{25} = \frac{80}{25} = \frac{16}{5}$.The distance from $C(\frac{4}{5}, \frac{8}{5})$ to the line $2x + 4y - 5 = 0$ is $d = \frac{|2(\frac{4}{5}) + 4(\frac{8}{5}) - 5|}{\sqrt{2^2 + 4^2}} = \frac{|\frac{8+32-25}{5}|}{\sqrt{20}} = \frac{15/5}{\sqrt{20}} = \frac{3}{\sqrt{20}}$.In $	riangle CPQ$,$CP^2 = d^2 + (PQ/2)^2$. Let $M$ be the midpoint of $PQ$. $CM = d = \frac{3}{\sqrt{20}}$.$PM^2 = r^2 - OM^2$. $OM$ is the distance from origin to $2x+4y=5$,$OM = \frac{|-5|}{\sqrt{20}} = \frac{5}{\sqrt{20}}$.$PM^2 = r^2 - \frac{25}{20} = r^2 - \frac{5}{4}$.$CP^2 = CM^2 + PM^2 \Rightarrow \frac{16}{5} = \frac{9}{20} + r^2 - \frac{5}{4} = r^2 + \frac{9-25}{20} = r^2 - \frac{16}{20} = r^2 - \frac{4}{5}$.$r^2 = \frac{16}{5} + \frac{4}{5} = \frac{20}{5} = 4 \Rightarrow r = 2$.

Similar Questions

The triangle $PQR$ is inscribed in the circle $x^2 + y^2 = 25$. If $Q$ and $R$ have coordinates $(3, 4)$ and $(-4, 3)$ respectively,then $\angle QPR$ is equal to

Suppose two perpendicular tangents can be drawn from the origin to the circle $x^2+y^2-6x-2py+17=0$,for some real $p$. Then,$|p|$ is equal to

If $(a, b)$ is the centre of the circle passing through the vertices of the triangle formed by $x+y=6, 2x+y=4$ and $x+2y=5$,then $(a, b)$ is

Let two circles $C_1$ and $C_2$ of radii $r_1 = 2$ and $r_2 = 4$ be tangent to each other at point $P$ and tangent to a common straight line (not passing through $P$) at points $Q$ and $R$ respectively. Then the value of $PQ^2 + QR^2 + RP^2$ is:

The value of $a$,such that the power of the point $(1, 6)$ with respect to the circle $x^2 + y^2 + 4x - 6y - a = 0$ is $-16$,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module