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10-1.Circle and System of Circles
hard
If line $ax + by = 0$ touches ${x^2} + {y^2} + 2x + 4y = 0$ and is a normal to the circle ${x^2} + {y^2} - 4x + 2y - 3 = 0$, then value of $(a,b)$ will be
A
$(2, 1)$
B
$(1, -2)$
C
$(1, 2)$
D
$(-1, 2)$
Solution
(c) As the line $ax + by = 0$ touches the circle ${x^2} + {y^2} + 2x + 4y = 0$,
distance of the centre $ (-1, -2)$ from the line = radius
==> $\left| {\frac{{ – a – 2b}}{{\sqrt {{a^2} + {b^2}} }}} \right| $
$= \sqrt {{{( – 1)}^2} + {{( – 2)}^2}} $
==> ${(a + 2b)^2} = 5({a^2} + {b^2})$
==> $4{a^2} – 4ab + {b^2} = 0$
==> ${(2a – b)^2} = 0$
$b = 2a$.
Next, $ax + by = 0$ is a normal to ${x^2} + {y^2} – 4x + 2y – 3 = 0$,
the centre $(2, -1)$ should lie on $ax + by = 0$
$2a – b = 0 \Rightarrow b = 2a$.
Hence $a = 1$, $b = 2$.
Standard 11
Mathematics
