The equation of a circle is operatorname{Re}( | vedclass

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

Question

(A) Given the equation of the circle: $\operatorname{Re}(z^{2})+2(\operatorname{Im}(z))^{2}+2 \operatorname{Re}(z)=0$.Since $z=x+iy$,we have $z^{2}=x^

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Given the equation of the circle: $\operatorname{Re}(z^{2})+2(\operatorname{Im}(z))^{2}+2 \operatorname{Re}(z)=0$.Since $z=x+iy$,we have $z^{2}=x^{2}-y^{2}+2ixy$,so $\operatorname{Re}(z^{2})=x^{2}-y^{2}$ and $\operatorname{Im}(z)=y$.The equation becomes $(x^{2}-y^{2})+2y^{2}+2x=0$,which simplifies to $x^{2}+y^{2}+2x=0$.The center of this circle is $(-1, 0)$.Given the parabola: $x^{2}-6x-y+13=0$.Rewriting as $(x-3)^{2}-9-y+13=0$,we get $(x-3)^{2}=y-4$.The vertex of the parabola is $(3, 4)$.The line passes through $(-1, 0)$ and $(3, 4)$.The slope $m = \frac{4-0}{3-(-1)} = \frac{4}{4} = 1$.The equation of the line is $y-0=1(x+1)$,which is $y=x+1$.The $y$-intercept is the value of $y$ when $x=0$,which is $1$.

The equation of a circle is $\operatorname{Re}(z^{2})+2(\operatorname{Im}(z))^{2}+2 \operatorname{Re}(z)=0$,where $z=x+iy$. $A$ line which passes through the center of the given circle and the vertex of the parabola $x^{2}-6x-y+13=0$ has a $y$-intercept equal to $.....$

Similar Questions

The equation of a circle touching the coordinate axes and the line $3x - 4y = 12$ is

If a circle touches both axes and lies below the line $4x + 3y = 6$ in the first quadrant,then the equation of the circle is:

If a circle with center $(0, 0)$ touches the line $5x + 12y = 1$,then its equation is:

The circle touching the $y$-axis at a distance $4$ units from the origin and cutting off an intercept $6$ from the $x$-axis is

If $(a, 1/a)$,$(b, 1/b)$,$(c, 1/c)$,and $(d, 1/d)$ are four distinct points on a circle of radius $4$ units,then $abcd$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module