The equation of a circle is Re (z2)+2(Im(z))2+2Re(z)=0, where z=x+ i y. A line which passes through the center of the given circle and the vertex of the parabola, x2-6 x-y+13=0, has y-intercept equal to.

Question

The equation of a circle is Re (z2)+2(Im(z))2+2Re(z)=0, where z=x+ i y. A line which passes through the center of the given circle and the vertex of the parabola, x2-6 x-y+13=0, has y-intercept equal to. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Equation of circle is (x2-y2)+2 y2+2 x=0 x2+y2+2 x=0 Centre: (-1,0) Parabola: x2-6 x-y+13=0 (x-3)2=y-4 Vertex: (3,4) Equation of line ≡ y-0=(4-0/3+1)(x+1) y=x+1 y-intercept =1

Q. The equation of a circle is $R e (z^{2}) + 2 (I m (z))^{2} + 2 R e (z) = 0$ , where $z = x + i y$ . A line which passes through the center of the given circle and the vertex of the parabola, $x^{2} - 6 x - y + 13 = 0$ , has $y$ -intercept equal to______.

Equation of circle is $(x^{2} - y^{2}) + 2 y^{2} + 2 x = 0$
$x^{2} + y^{2} + 2 x = 0$
Centre: $(- 1, 0)$
Parabola $: x^{2} - 6 x - y + 13 = 0$
$(x - 3)^{2} = y - 4$
Vertex: $(3, 4)$
Equation of line $\equiv y - 0 = \frac{4 - 0}{3 + 1} (x + 1)$
$y = x + 1$
$y$ -intercept $= 1$

Q. The equation of a circle is Re(z2)+2(Im(z))2+2Re(z)=0, where z=x+iy. A line which passes through the center of the given circle and the vertex of the parabola, x2−6x−y+13=0, has y-intercept equal to______.

Equation of circle is (x2−y2)+2y2+2x=0 x2+y2+2x=0 Centre: (−1,0) Parabola :x2−6x−y+13=0 (x−3)2=y−4 Vertex: (3,4) Equation of line ≡y−0=3+14−0​(x+1) y=x+1 y-intercept =1

Q. The equation of a circle is $R e (z^{2}) + 2 (I m (z))^{2} + 2 R e (z) = 0$ , where $z = x + i y$ . A line which passes through the center of the given circle and the vertex of the parabola, $x^{2} - 6 x - y + 13 = 0$ , has $y$ -intercept equal to______.

Equation of circle is $(x^{2} - y^{2}) + 2 y^{2} + 2 x = 0$
$x^{2} + y^{2} + 2 x = 0$
Centre: $(- 1, 0)$
Parabola $: x^{2} - 6 x - y + 13 = 0$
$(x - 3)^{2} = y - 4$
Vertex: $(3, 4)$
Equation of line $\equiv y - 0 = \frac{4 - 0}{3 + 1} (x + 1)$
$y = x + 1$
$y$ -intercept $= 1$