If a line along a chord of the circle 4x^{2}+ | vedclass

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

Question

(D) The equation of the circle is $4x^{2}+4y^{2}+120x+675=0$,which simplifies to $x^{2}+y^{2}+30x+\frac{675}{4}=0$.Completing the square: $(x+15)^{2}+

Vedclass · Accepted Answer

$3\sqrt{5}$

Vedclass · Answer

(D) The equation of the circle is $4x^{2}+4y^{2}+120x+675=0$,which simplifies to $x^{2}+y^{2}+30x+\frac{675}{4}=0$.Completing the square: $(x+15)^{2}+y^{2} = 225 - \frac{675}{4} = \frac{900-675}{4} = \frac{225}{4}$.So,the center is $(-15, 0)$ and the radius $R = \sqrt{\frac{225}{4}} = \frac{15}{2}$.The equation of a line passing through $(-30, 0)$ is $y = m(x+30)$,or $mx - y + 30m = 0$.This line is tangent to the parabola $y^{2}=30x$. The condition for tangency $y=mx+c$ to $y^{2}=4ax$ is $c = \frac{a}{m}$.Here $4a=30 \Rightarrow a = \frac{15}{2}$. The line is $y = mx + 30m$,so $c = 30m$.Thus,$30m = \frac{15/2}{m}$ $\Rightarrow 60m^{2} = 15$ $\Rightarrow m^{2} = \frac{1}{4}$ $\Rightarrow m = \pm \frac{1}{2}$.Taking $m = \frac{1}{2}$,the line is $y = \frac{1}{2}(x+30) \Rightarrow x - 2y + 30 = 0$.The perpendicular distance $P$ from the center $(-15, 0)$ to the line $x - 2y + 30 = 0$ is:$P = \frac{|-15 - 2(0) + 30|}{\sqrt{1^{2} + (-2)^{2}}} = \frac{15}{\sqrt{5}} = 3\sqrt{5}$.The length of the chord is $2\sqrt{R^{2}-P^{2}} = 2\sqrt{(\frac{15}{2})^{2} - (3\sqrt{5})^{2}} = 2\sqrt{\frac{225}{4} - 45} = 2\sqrt{\frac{225-180}{4}} = 2\sqrt{\frac{45}{4}} = \sqrt{45} = 3\sqrt{5}$.

If a line along a chord of the circle $4x^{2}+4y^{2}+120x+675=0$ passes through the point $(-30, 0)$ and is tangent to the parabola $y^{2}=30x$,then the length of this chord is:

Similar Questions

$B$ and $C$ are fixed points having coordinates $(3, 0)$ and $(-3, 0)$ respectively. If the vertical angle $\angle BAC$ is $90^o$,then the locus of the centroid of the $\Delta ABC$ has the equation:

Let $AB$ be a chord of the circle $x^2 + y^2 = r^2$ subtending a right angle at the centre. Then the locus of the centroid of the $\Delta PAB$ as $P$ moves on the circle is

Two tangents $PQ$ and $PR$ are drawn to the circle $x^2 + y^2 - 2x - 4y - 20 = 0$ from the point $P(16, 7)$. If the centre of the circle is $C$,then the area of quadrilateral $PQCR$ is ............ $sq. \text{ units}$.

If the image of the point $(-4, 5)$ in the line $x + 2y = 2$ lies on the circle $(x + 4)^2 + (y - 3)^2 = r^2$,then $r$ is equal to:

Consider the equations of the circles:
$S_1 : x^2 + y^2 + 24x - 10y + a = 0$
$S_2 : x^2 + y^2 = 36$
Which of the following statements is not correct?

Vedclass Test Series

Exam Paper Generator

Online Exam Module