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 power of a point $(4, 2)$ with respect to the circle $x^2 + y^2 - 2\alpha x + 6y + \alpha^2 - 16 = 0$ is $9$,then the sum of the lengths of all possible intercepts made by such circles on the coordinate axes is:

$A$ line $y=mx+1$ intersects the circle $(x-3)^2+(y+2)^2=25$ at the points $P$ and $Q$. If the midpoint of the line segment $PQ$ has $x$-coordinate $-\frac{3}{5}$,then which one of the following options is correct?

In a $\triangle ABC$,a point $D$ is chosen on $BC$ such that $BD:DC = 2:5$. Let $P$ be a point on the circumcircle of $\triangle ABC$ such that $\angle PDB = \angle BAC$. Then $PD:PC$ is:

For how many values of $p$ does the circle $x^2+y^2+2x+4y-p=0$ and the coordinate axes have exactly three common points?

Two circles of radii $4 \text{ cm}$ and $1 \text{ cm}$ touch each other externally and $\theta$ is the angle contained by their direct common tangents. Then $\sin \theta =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module