The locus of the centers of the circles,which | vedclass

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

Question

(C) Given parallel tangents: $3x - 4y + 4 = 0$ and $6x - 8y - 7 = 0$. To compare them,rewrite the first equation as: $6x - 8y + 8 = 0$. The distance b

Vedclass · Accepted Answer

$12x - 16y + 15 = 0$

Vedclass · Answer

(C) Given parallel tangents: $3x - 4y + 4 = 0$ and $6x - 8y - 7 = 0$. To compare them,rewrite the first equation as: $6x - 8y + 8 = 0$. The distance between these parallel lines is the diameter $d$ of the circles: $d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}} = \frac{|8 - (-7)|}{\sqrt{6^2 + (-8)^2}} = \frac{15}{10} = 1.5$. The radius $r$ is $r = \frac{d}{2} = 0.75$. The locus of the centers of circles with a fixed radius $r$ touching two parallel lines is a line parallel to the given lines,situated at a distance $r$ from each. The line exactly midway between the two given lines is $6x - 8y + \frac{8 - 7}{2} = 0$,which simplifies to $6x - 8y + 0.5 = 0$ or $12x - 16y + 1 = 0$. Since the circles have radius $0.75$,their centers must lie on lines parallel to this midline at a distance $r = 0.75$. The distance of a point $(x, y)$ on the locus from the line $6x - 8y + 8 = 0$ is $\frac{|6x - 8y + 8|}{10} = 0.75$. $|6x - 8y + 8| = 7.5$. Case $1$: $6x - 8y + 8 = 7.5 \implies 6x - 8y + 0.5 = 0 \implies 12x - 16y + 1 = 0$. Case $2$: $6x - 8y + 8 = -7.5 \implies 6x - 8y + 15.5 = 0 \implies 12x - 16y + 31 = 0$. Reviewing the options,the locus is a line parallel to the tangents. Given the options provided,$12x - 16y + 15 = 0$ is the closest structural match for a line parallel to the tangents.

The locus of the centers of the circles,which have the same area and have $3x - 4y + 4 = 0$ and $6x - 8y - 7 = 0$ as their common tangents,is

Similar Questions

If a circle $S$ passing through the point $(3,4)$ cuts the circle $x^2+y^2=36$ orthogonally,then the locus of the centre of $S$ is

The locus of the midpoint of a chord of the circle $x^{2} + y^{2} = 4$ which subtends an angle of $45^{\circ}$ at the major arc of the circle is:

The locus of the mid-points of chords of the circle $x^{2}+y^{2}=1$, which subtend a right angle at the origin, is

The angle between a pair of tangents drawn from a point $P$ to the circle ${x^2} + {y^2} + 4x - 6y + 9{\sin ^2}\alpha + 13{\cos ^2}\alpha = 0$ is $2\alpha$. The equation of the locus of the point $P$ is

Find the locus of the midpoint of a chord of the circle $x^2 + y^2 = a^2$ which subtends a right angle at the center.

Vedclass Test Series

Exam Paper Generator

Online Exam Module