If alpha neq -4 and (2, alpha) is the mid-poi | vedclass

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

Question

(A) The equation of the circle is $x^2+y^2-4x+8y+6=0$. The center is $(2, -4)$ and the radius is $r = \sqrt{2^2+(-4)^2-6} = \sqrt{4+16-6} = \sqrt{14}$

Vedclass · Accepted Answer

$(-4-\sqrt{14}, -4+\sqrt{14})$

Vedclass · Answer

(A) The equation of the circle is $x^2+y^2-4x+8y+6=0$. The center is $(2, -4)$ and the radius is $r = \sqrt{2^2+(-4)^2-6} = \sqrt{4+16-6} = \sqrt{14}$.Since $(2, \alpha)$ is the mid-point of a chord,it must lie inside the circle. Substituting $(2, \alpha)$ into the circle equation gives $2^2+\alpha^2-4(2)+8\alpha+6 Solving $\alpha^2+8\alpha+2=0$ gives $\alpha = -4 \pm \sqrt{14}$. Thus,$\alpha \in (-4-\sqrt{14}, -4+\sqrt{14})$.The chord passes through $(2, \alpha)$ and is perpendicular to the radius connecting $(2, -4)$ and $(2, \alpha)$. The slope of the radius is undefined (vertical line $x=2$),so the chord is a horizontal line $y=\alpha$.The $y$-intercept of the line $y=\alpha$ is simply $\alpha$. Therefore,the $y$-intercept lies in the interval $(-4-\sqrt{14}, -4+\sqrt{14})$.

If $\alpha \neq -4$ and $(2, \alpha)$ is the mid-point of a chord of the circle $x^2+y^2-4x+8y+6=0$,then the values of the $y$-intercept of the chord lie in the interval

Similar Questions

If the radius of the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is $r$,then it will touch both the axes if:

Let $A(2, 3), B(4, 5)$ be two points and let $C = (x, y)$ be a point such that $(x - 2)(x - 4) + (y - 3)(y - 5) = 0$. If the area of $\Delta ABC = \sqrt{2}$ sq. unit,then the maximum number of positions of $C$ in the $xy$ plane is

Suppose the tangents drawn to the circle $x^2+y^2-6x-4y-11=0$ from $P(1,8)$ touch the circle at $A$ and $B$. Then the centre of the circle passing through $P, A$ and $B$ is

The centre of the smallest circle touching the circles $x^2 + y^2 - 2y - 3 = 0$ and $x^2 + y^2 - 8x - 18y + 93 = 0$ is:

If a circle $S$ passing through the points $A(1, 2)$ and $B(2, 1)$ has its centre $C$ located in the third quadrant at a distance of $\frac{7}{\sqrt{2}}$ units from the line $AB$,then the point $P(1, -2)$:

Vedclass Test Series

Exam Paper Generator

Online Exam Module