If the lines $2 x + 3 y + 19 = 0$ and $9 x + 6 y - 17 = 0$ intersect the axes at four points, then the equation of the circle through these points is

18 x^{2} + 18 y^{2} + 137 x + 63 y + 323 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

18 x^{2} + 18 y^{2} + 137 x + 63 y - 323 = 0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

18 x^{2} + 18 y^{2} - 137 x + 63 y - 323 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

18 x^{2} + 18 y^{2} + 137 x - 63 y - 323 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $18 x^{2} + 18 y^{2} + 137 x + 63 y - 323 = 0$
$a_{1} = 2, b_{1} = 3, a_{2} = 9, b_{2} = 6$
$a_{1} a_{2} = b_{1} b_{2} \Rightarrow 2 \times 9 = 3 \times 6$
The desired circle is given by $L_{1} L_{2} = 0$ ignoring $x y$ term.
$(2 x + 3 y + 19) (9 x + 6 y - 17) = 0$
On simplifying, we get
$\Rightarrow 18 x^{2} + 18 y^{2} + 137 x + 63 y - 323 = 0$

Suggest Corrections

Similar questions

2 x + 3 y + 19 = 0

9 x + 6 y - 17 = 0

8 x^{2} - 24 x y + 18 y^{2} - 6 x + 9 y - 5 = 0

9 x^{2} - 6 x y + y^{2} + 18 x - 6 y + 8 = 0

x^{2} + 2 \sqrt{3} x y + 3 y^{2} - 3 x - 3 \sqrt{3} y - 4 = 0

2 x + 3 y + 19 = 0

9 x + 6 y - 17 = 0

t

8

18 x^{2} + 9 x - t = 0

y = 0, 37 x - 36 y + 37 \times 36 = 0

64 x - 63 y + 64 \times 63 = 0

(a, b),

a - b

Join BYJU'S Learning Program