The difference between the radii of the largest and the smallest circles which have their centers on the circumference of the circles $x^{2} + y^{2} + 2 x + 4 y - 4 = 0$ and pass through the point $(a, b)$ lying outside the given circle is

3

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

6

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

5

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

None of these

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

Open in App

Solution

The correct option is B $6$
The given circle is ${(x + 1)}^{2} + {(y + 2)}^{2} = 9,$ having centre at $(- 1, - 2)$ and radius $3.$
Clearly, from the figure, the points on the circle which are nearest and farthest to the point $P (a, b)$ are $Q$ and $R,$ respectively.
Thus, the circle centred at $Q$ having $P Q$ will be the smallest required circle, while the circle centred at $Q$ having radius $P R$ will be the largest re-quired circle.
Hence, difference between their radii $= P R - P Q = Q R = 6.$

Suggest Corrections

Similar questions

x^{2} + y^{2} + 2 x + 4 y - 4 = 0

(a, b)

r_{1}

r_{2}

(- 4, 1)

x^{2} + y^{2} + 2 x + 4 y - 4 = 0.

\frac{r_{1}}{r_{2}} = a + b \sqrt{2,}

a + b

The equation of the circle passing through the point (1, 1) and having two diameters along the pair of lines
$x^{2} - y^{2} - 2 x + 4 y - 3 = 0,$ is

x + 2 y - 3 = 0

x^{2} + y^{2} - 2 x - 4 y + 1 = 0

x^{2} + y^{2} - 4 x - 2 y + 4 = 0

Join BYJU'S Learning Program