Question

Match the entries of List-A and List-B.
List-A List-B

Answer 1

(a) Any line through (1, 7) is
y - 7 = m(x - 1) or mx - y + (7 - m) = 0
Circle is (1, 2), 3. Apply p = r for tangency.
$\frac{m-2+7-m}{\sqrt{1+m^2}}=3$
or $25=9+9m^2$
$\therefore m=\pm \frac{4}{3}$

(b) Let the circle be
$x^2+y^2+2gx+2fy+c=0$.
It touches y-axis i.e.x = 0 at (0, 3)
$\therefore y^2+2fy+c=(y-3{)}^2=y^2-6y+9$
Comparing f = -3, c=9.
Intercept of x-axis is
$I_x=2\sqrt{g^2-c}=8\therefore g^2-9=16$
or $g^2=16+9=25org=\pm 5$.
Hence the required equation is
$x^2+y^2\pm 10x-6y+9=0$
(c) $s_1+\lambda S_2=0$. Reduce to standard form and find its centre which lies on $y=x⟹\lambda =4/3$
(d) $S_1-S_2=0⟹x-y=0$ is a diameter
$\therefore$ Circle is $S+\lambda P=0$
$(x^2+y^2+2x)+\lambda (x-y)=0$
$\therefore \lambda =-1$
(e) Common tangent is given by $S_1-S_2=0$ or ax - by = 0. The condition of tangency p = r from (a, 0), $\sqrt{a^2-c^2}$ or $(0,b)\sqrt{b^2-c^2}$ gives
$\frac{a^2}{\sqrt{a^2+b^2}}=\sqrt{a^2-c^2}$
or $a^4=(a^2+b^2)(a^2-c^2)or\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}$
(f) $(C_1{C}_2{)}^2=r_1^2+r_2^{2}or18=a^2+a^2$