SATHEE: Circle 5 Question 10

Circle 5 Question 10

10. The equation of the locus of the mid-points of the chords of the circle $4 x^{2} + 4 y^{2} - 12 x + 4 y + 1 = 0$ that subtend an angle of $2 π / 3$ at its centre is … .

(1993, 2M)

Show Answer

Answer:

Correct Answer: 10. $16 x^{2} + 16 y^{2} - 48 x + 16 y + 31 = 0$

Solution:

Given, $4 x^{2} + 4 y^{2} - 12 x + 4 y + 1 = 0$

$\Rightarrow x^{2} + y^{2} - 3 x + y + 1 / 4 = 0$ whose centre is $\frac{3}{2}, - \frac{1}{2}$ and radius

$\begin{aligned} = \sqrt{{\frac{3}{2}}^{2} + - {\frac{1}{2}}^{2} - \frac{1}{4}} = \sqrt{\frac{9}{4} + \frac{1}{4} - \frac{1}{4}} \\ = \frac{3}{2} \end{aligned}$

Again, let $S$ be a circle with centre at $C$ and $A B$ is given chord and $A D$ subtend angle $2 π / 3$ at the centre and $D$ be the mid point of $A B$ and let its coordinates are $(h, k)$ .

Now, $∠ D C A = \frac{1}{2} (∠ B C A) = \frac{1}{2} \cdot \frac{2 π}{3} = \frac{π}{3}$

Using sine rule in $△ A D C$ ,

$\frac{D A}{\sin π / 3} = \frac{C A}{\sin π / 2}$

$\Rightarrow D A = C A \sin π / 3 = \frac{3}{2} \cdot \frac{\sqrt{3}}{2}$

Now, in $△ A C D$

$\begin{array}{ll} C D^{2} = C A^{2} - A D^{2} = \frac{9}{4} - \frac{27}{16} = \frac{9}{16} \\ But & C D^{2} = (h - 3 / 2)^{2} + (k + 1 / 2)^{2} \\ \Rightarrow & (h - 3 / 2)^{2} + (k + 1 / 2)^{2} = \frac{9}{16} \end{array}$

Hence, locus of a point is

$\begin{aligned} x - {\frac{3}{2}}^{2} + y + {\frac{1}{2}}^{2} & = \frac{9}{16} \\ \Rightarrow 16 x^{2} + 16 y^{2} - 48 x + 16 y + 31 & = 0 \end{aligned}$

Circle 5 Question 10

10. The equation of the locus of the mid-points of the chords of the circle $4 x^{2} + 4 y^{2} - 12 x + 4 y + 1 = 0$ that subtend an angle of $2 π / 3$ at its centre is … .

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Circle 5 Question 10

10. The equation of the locus of the mid-points of the chords of the circle 4x2+4y2−12x+4y+1=0 that subtend an angle of 2π/3 at its centre is … .

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu

10. The equation of the locus of the mid-points of the chords of the circle $4 x^{2} + 4 y^{2} - 12 x + 4 y + 1 = 0$ that subtend an angle of $2 π / 3$ at its centre is … .