From a point P(0, b),two tangents are drawn t | vedclass

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

Question

(C) The equation of the pair of tangents from point $P(0, b)$ to the circle $x^2+y^2=16$ is given by $SS_1 = T^2$,which is $(x^2+y^2-16)(b^2-16) = (0

Vedclass · Accepted Answer

$x^2+y^2=32$

Vedclass · Answer

(C) The equation of the pair of tangents from point $P(0, b)$ to the circle $x^2+y^2=16$ is given by $SS_1 = T^2$,which is $(x^2+y^2-16)(b^2-16) = (0 \cdot x + b \cdot y - 16)^2$,or $(x^2+y^2-16)(b^2-16) = (by-16)^2$.For points $A$ and $B$ on the $X$-axis,we set $y=0$ in the equation:$(x^2-16)(b^2-16) = (-16)^2 = 256$$x^2-16 = \frac{256}{b^2-16}$$x^2 = 16 + \frac{256}{b^2-16} = \frac{16b^2-256+256}{b^2-16} = \frac{16b^2}{b^2-16}$$x = \pm \frac{4b}{\sqrt{b^2-16}}$Thus,the coordinates of $A$ and $B$ are $(\pm \frac{4b}{\sqrt{b^2-16}}, 0)$.The area of $	riangle PAB$ is $\Delta = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes \left( \frac{8b}{\sqrt{b^2-16}} ight) 	imes |b| = \frac{4b^2}{\sqrt{b^2-16}}$.To minimize the area,we differentiate $\Delta$ with respect to $b$ and set it to $0$:$\frac{d\Delta}{db} = 4 \left[ \frac{2b\sqrt{b^2-16} - b^2 \cdot \frac{2b}{2\sqrt{b^2-16}}}{b^2-16} ight] = 0$$2b(b^2-16) - b^3 = 0 \Rightarrow 2b^3 - 32b - b^3 = 0 \Rightarrow b^3 = 32b$Since $b 
eq 0$,$b^2 = 32$.For $b^2=32$,the $x$-coordinates are $\pm \frac{4\sqrt{32}}{\sqrt{32-16}} = \pm \frac{4 \cdot 4\sqrt{2}}{4} = \pm 4\sqrt{2}$.The vertices are $P(0, 4\sqrt{2})$,$A(4\sqrt{2}, 0)$,and $B(-4\sqrt{2}, 0)$.Since $	riangle PAB$ is a right-angled triangle with the right angle at $P$ (or by observing the symmetry),the circumcircle has the segment $AB$ as its diameter if it were a right triangle at $P$,but here the vertices are $(0, b), (x_0, 0), (-x_0, 0)$. The circumcenter $(0, y_c)$ satisfies $y_c^2 + x_0^2 = (b-y_c)^2$. Substituting $x_0^2 = \frac{16b^2}{b^2-16} = \frac{16(32)}{16} = 32$ and $b^2=32$,we find the circumcircle equation $x^2 + (y-y_c)^2 = R^2$.

From a point $P(0, b)$,two tangents are drawn to the circle $x^2+y^2=16$. These two tangents intersect the $X$-axis at two points $A$ and $B$. If the area of $\triangle PAB$ is minimum,then the equation of its circumcircle is

Similar Questions

Consider the two curves $C_1: y^2=4x$ and $C_2: x^2+y^2-6x+1=0$. Then,

If two chords,each bisected by the $x$-axis,can be drawn to the circle $2(x^2 + y^2) - 2ax - by = 0$ $(a \ne 0, b \ne 0)$ from the point $P(a, b/2)$,then:

If the tangent at the point $\left(4 \cos 2 \theta, \frac{16}{\sqrt{11}} \sin 2 \theta\right)$ on the ellipse $16 x^2+11 y^2=256$ touches the circle $x^2+y^2-2 x=15$,then $\theta=$

$A$ focal chord to $y^2 = 16x$ is a tangent to $(x - 6)^2 + y^2 = 2$. Then the possible values of the slope of this chord are:

Let the line $x-y+1=0$ intersect the circle $x^2+y^2+2x+2y+1=0$ at two points $A$ and $B$. If $AB$ is the diameter of the circle $x^2+y^2+2gx+2fy+c=0$,then $g+f=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module