Let PQ and MN be two straight lines touching the circle x 2 + y 2 - 4 x - 6 y - 3 = 0 at the points A and B respectively. Let O be the centre of the circle and angle A O B = 3 Then the locus of the point of intersection of the lines PQ and MN is: [20

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Q.
Let PQ and MN be two straight lines touching the circle $x^{2} + y^{2} - 4 x - 6 y - 3 = 0$ at the points A and B respectively. Let O be the centre of the circle and angle $∠ A O B = \frac{π}{3}$ Then the locus of the point of intersection of the lines PQ and MN is: [2026]

1 $3 (x^{2} + y^{2}) - 12 x - 18 y - 25 = 0$

2 $3 (x^{2} + y^{2}) - 18 x - 12 y + 25 = 0$

3 $x^{2} + y^{2} - 12 x - 18 y - 25 = 0$

4 $x^{2} + y^{2} - 18 x - 12 y - 25 = 0$

Ans.
(1)

Given Circle

$x^{2} + y^{2} - 4 x - 6 y - 3 = 0$

$C (2, 3) and r = 4$

$\cos 30 ° = \frac{r}{O R} = \frac{4}{O R}$

$\Rightarrow O R = \frac{8}{\sqrt{3}}$

Now

$O R^{2} = {(h - 2)}^{2} + {(k - 3)}^{2}$

$\Rightarrow 3 (x^{2} + y^{2}) - 12 x - 18 y - 25 = 0$

Solution

Q.
Let PQ and MN be two straight lines touching the circle $x^{2} + y^{2} - 4 x - 6 y - 3 = 0$ at the points A and B respectively. Let O be the centre of the circle and angle $∠ A O B = \frac{π}{3}$ Then the locus of the point of intersection of the lines PQ and MN is: [2026]

1 $3 (x^{2} + y^{2}) - 12 x - 18 y - 25 = 0$

2 $3 (x^{2} + y^{2}) - 18 x - 12 y + 25 = 0$

3 $x^{2} + y^{2} - 12 x - 18 y - 25 = 0$

4 $x^{2} + y^{2} - 18 x - 12 y - 25 = 0$

Ans.
(1)

Given Circle

$x^{2} + y^{2} - 4 x - 6 y - 3 = 0$

$C (2, 3) and r = 4$

$\cos 30 ° = \frac{r}{O R} = \frac{4}{O R}$

$\Rightarrow O R = \frac{8}{\sqrt{3}}$

Now

$O R^{2} = {(h - 2)}^{2} + {(k - 3)}^{2}$

$\Rightarrow 3 (x^{2} + y^{2}) - 12 x - 18 y - 25 = 0$

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Solution

Q. Let PQ and MN be two straight lines touching the circle x2+y2-4x-6y-3=0 at the points A and B respectively. Let O be the centre of the circle and angle ∠AOB=π3 Then the locus of the point of intersection of the lines PQ and MN is: [2026]

1 3(x2+y2)-12x-18y-25=0

2 3(x2+y2)-18x-12y+25=0

3 x2+y2-12x-18y-25=0

4 x2+y2-18x-12y-25=0

Ans. (1) Given Circle x2+y2-4x-6y-3=0 C(2,3) and r=4 cos30°=rOR=4OR ⇒OR=83 Now OR2=(h-2)2+(k-3)2 ⇒3(x2+y2)-12x-18y-25=0

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Q.
Let PQ and MN be two straight lines touching the circle $x^{2} + y^{2} - 4 x - 6 y - 3 = 0$ at the points A and B respectively. Let O be the centre of the circle and angle $∠ A O B = \frac{π}{3}$ Then the locus of the point of intersection of the lines PQ and MN is: [2026]

1 $3 (x^{2} + y^{2}) - 12 x - 18 y - 25 = 0$

2 $3 (x^{2} + y^{2}) - 18 x - 12 y + 25 = 0$

3 $x^{2} + y^{2} - 12 x - 18 y - 25 = 0$

4 $x^{2} + y^{2} - 18 x - 12 y - 25 = 0$

Ans.
(1)

Given Circle

$x^{2} + y^{2} - 4 x - 6 y - 3 = 0$

$C (2, 3) and r = 4$

$\cos 30 ° = \frac{r}{O R} = \frac{4}{O R}$

$\Rightarrow O R = \frac{8}{\sqrt{3}}$

Now

$O R^{2} = {(h - 2)}^{2} + {(k - 3)}^{2}$

$\Rightarrow 3 (x^{2} + y^{2}) - 12 x - 18 y - 25 = 0$