Let S be the circle in the xy-plane defined b | vedclass

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

Question

(A,D) $(1)$ The circle is $x^2+y^2=4$. The point $P_0(1,1)$ lies inside the circle.For chord $E_1E_2$ parallel to the $x$-axis,$y=1$. Substituting int

Vedclass · Accepted Answer

Vedclass · Answer

(A,D) $(1)$ The circle is $x^2+y^2=4$. The point $P_0(1,1)$ lies inside the circle.For chord $E_1E_2$ parallel to the $x$-axis,$y=1$. Substituting into the circle equation: $x^2+1=4 \implies x^2=3 \implies x=\pm\sqrt{3}$. Thus $E_1(-\sqrt{3}, 1)$ and $E_2(\sqrt{3}, 1)$.The tangent at $(x_1, y_1)$ is $xx_1+yy_1=4$. Tangents at $E_1, E_2$ are $-x\sqrt{3}+y=4$ and $x\sqrt{3}+y=4$. Solving these gives $E_3(0, 4)$.For chord $F_1F_2$ parallel to the $y$-axis,$x=1$. Substituting into the circle equation: $1+y^2=4 \implies y^2=3 \implies y=\pm\sqrt{3}$. Thus $F_1(1, \sqrt{3})$ and $F_2(1, -\sqrt{3})$.Tangents at $F_1, F_2$ are $x+y\sqrt{3}=4$ and $x-y\sqrt{3}=4$. Solving these gives $F_3(4, 0)$.For chord $G_1G_2$ with slope $-1$ passing through $(1,1)$,the line is $y-1=-1(x-1) \implies x+y=2$. Intersection with $x^2+y^2=4$: $x^2+(2-x)^2=4 \implies 2x^2-4x=0 \implies x=0, 2$. Thus $G_1(0, 2)$ and $G_2(2, 0)$.Tangents at $G_1(0, 2)$ is $y=2$. Tangent at $G_2(2, 0)$ is $x=2$. Intersection $G_3(2, 2)$.Points $E_3(0, 4), F_3(4, 0), G_3(2, 2)$ all satisfy $x+y=4$. Correct option is $(A)$.$(2)$ Let $P(2\cos	heta, 2\sin	heta)$. The tangent is $x\cos	heta+y\sin	heta=2$. Intercepts are $M(2/\cos	heta, 0)$ and $N(0, 2/\sin	heta)$.Mid-point $(h, k) = (1/\cos	heta, 1/\sin	heta)$.Thus $\cos	heta=1/h$ and $\sin	heta=1/k$. Since $\cos^2	heta+\sin^2	heta=1$,we have $1/h^2+1/k^2=1 \implies h^2+k^2=h^2k^2$.Locus is $x^2+y^2=x^2y^2$. Correct option is $(D)$.

Similar Questions

The equation of the common tangent to the parabola $y^2=8x$ and the circle $x^2+y^2=2$ is $ax+by+2=0$. If $-\frac{a}{b} > 0$,then $3a^2+2b+1=$

If the equation of the circumcircle of the triangle formed by the lines $L_1 \equiv x+y=0$,$L_2 \equiv 2x+y-1=0$,and $L_3 \equiv x-3y+2=0$ is $\lambda_1 L_1 L_2 + \lambda_2 L_2 L_3 + \lambda_3 L_3 L_1 = 0$,then find the value of $\frac{7 \lambda_1}{\lambda_2} + \frac{\lambda_3}{\lambda_1}$.

The sum of diameters of the circles that touch $(i)$ the parabola $75x^2 = 64(5y - 3)$ at the point $\left(\frac{8}{5}, \frac{6}{5}\right)$ and $(ii)$ the $y$-axis,is equal to $......$

If the squares of the lengths of the tangents drawn from a point $P$ to the circles $x^{2} + y^{2} = a^2$,$x^2 + y^{2} = b^2$,and $x^{2} + y^{2} = c^{2}$ are in arithmetic progression,then:

The locus of the midpoints of the chords of the circle $x^2 + y^2 + 4x - 6y - 12 = 0$ which subtend an angle of $\frac{\pi}{3}$ radians at its circumference is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module