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 sum of the squares of the lengths of the chords intercepted on the circle,$x^2 + y^2 = 16$,by the lines,$x + y = n$,$n \in N$,where $N$ is the set of all natural numbers,is

If the angle between the circles $x^2+y^2-2x-4y+c=0$ and $x^2+y^2-4x-2y+4=0$ is $60^{\circ}$,then $c=$

The line $2x + y = 4$ intersects the $X$ and $Y$ axes at points $A$ and $B$ respectively. Let $C(p, q)$ be a point such that the lines $AC$ and $BC$ are perpendicular. Find the distance of point $C$ from the midpoint of segment $AB$.

The point $P(10, 7)$ lies outside the circle $x^2 + y^2 - 4x - 2y - 20 = 0$. The greatest distance of $P$ from the circle is

Let the circle with centre at origin pass through the vertices of an equilateral triangle $ABC$. If $A = (2, 4)$,then the length of the median through $A$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module