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10-1.Circle and System of Circles
normal
If the line $3x -4y -k = 0 (k > 0)$ touches the circle $x^2 + y^2 -4x -8y -5 = 0$ at $(a, b)$ then $k + a + b$ is equal to :-
A
$20$
B
$22$
C
$-30$
D
$-28$
Solution
since, the given line touches the given circle, the length of the perpendicular from the centre $(2,4)$ of the circle to the line $3 x-4 y-k=0$ is equal
to the radius $\sqrt{4+16+5}=5$ of the circle.
$\therefore \frac{3 \times 2-4 \times 4-k}{\sqrt{9+16}}=\pm 5$
$\Rightarrow \quad k=15$ $[\because k>0]$
hence equation of tangent is
$3 x-4 y-15=0$ …….$(1)$
Let equation of normal to circle
$4 x+3 y=\lambda$
It passes through centre $(2,4)$
$\Rightarrow \lambda=20$
hence equation of normal is
$4 x+3 y=20$ ………$(2)$
Solve $( 1)$ and $(2)$
$a=5, \quad b=0$
$k+a+b=15+5+0=20$
Standard 11
Mathematics