Let the line x-y+1=0 intersect the circle x^2 | vedclass

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

Question

(C) Given the line $x-y+1=0$,we have $y=x+1$. Substituting this into the circle equation $x^2+y^2+2x+2y+1=0$: $x^2+(x+1)^2+2x+2(x+1)+1=0$ $x^2+x^2+2x+

Vedclass · Accepted Answer

$c$

Vedclass · Answer

(C) Given the line $x-y+1=0$,we have $y=x+1$. Substituting this into the circle equation $x^2+y^2+2x+2y+1=0$: $x^2+(x+1)^2+2x+2(x+1)+1=0$ $x^2+x^2+2x+1+2x+2x+2+1=0$ $2x^2+6x+4=0 \Rightarrow x^2+3x+2=0$ $(x+1)(x+2)=0$,so $x=-1$ or $x=-2$. For $x=-1$,$y=0$. For $x=-2$,$y=-1$. Thus,the points are $A(-1, 0)$ and $B(-2, -1)$. The equation of a circle with diameter $AB$ is $(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$. $(x+1)(x+2)+(y-0)(y+1)=0$ $x^2+3x+2+y^2+y=0$ $x^2+y^2+3x+y+2=0$. Comparing this with $x^2+y^2+2gx+2fy+c=0$,we get $2g=3 \Rightarrow g=3/2$,$2f=1 \Rightarrow f=1/2$,and $c=2$. Then $g+f = 3/2 + 1/2 = 2$. Since $c=2$,we have $g+f=c$.

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 $g+f=$

Similar Questions

If the point $(1,4)$ lies inside the circle $x^2+y^2-6x-10y+p=0$ and the circle does not touch or intersect the coordinate axes,then

The locus of the centre of the circle which touches the circle ${x^2} + {(y - 1)^2} = 1$ externally and also touches the $x$-axis is

The locus of the midpoints of the chords of the circle $x^2 + y^2 - ax - by = 0$ which subtend a right angle at $\left( \frac{a}{2}, \frac{b}{2} \right)$ is:

Let $a$ and $b$ be two non-zero real numbers. The equation $(ax^2 + by^2 + c)(x^2 - 5xy + 6y^2) = 0$ represents:

$ax - y + c = 0$ is the equation of the common tangent to the parabola $y^2 = 8\sqrt{5}x$ and the circle $x^2 + y^2 = 1$. If this tangent makes an acute angle with the positive $X$-axis,then $a^2c^2 =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module