If the straight line ax + by = 2 where a, b n | vedclass

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(C) Given the first circle $x^2 + y^2 - 2x - 3 = 0$,the center is $(1, 0)$ and the radius $r_1 = \sqrt{1^2 + 0^2 - (-3)} = 2$.Given the second circle

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$-\frac{4}{3}, 1$

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(C) Given the first circle $x^2 + y^2 - 2x - 3 = 0$,the center is $(1, 0)$ and the radius $r_1 = \sqrt{1^2 + 0^2 - (-3)} = 2$.Given the second circle $x^2 + y^2 - 4y - 6 = 0$,the center is $(0, 2)$ and the radius $r_2 = \sqrt{0^2 + 2^2 - (-6)} = \sqrt{10}$.Since the line $ax + by = 2$ touches the first circle,the perpendicular distance from the center $(1, 0)$ to the line equals the radius:$\frac{|a(1) + b(0) - 2|}{\sqrt{a^2 + b^2}} = 2 \implies |a - 2| = 2\sqrt{a^2 + b^2} \dots (i)$.Since the line is normal to the second circle,it must pass through its center $(0, 2)$:$a(0) + b(2) = 2 \implies 2b = 2 \implies b = 1$.Substituting $b = 1$ into equation $(i)$:$|a - 2| = 2\sqrt{a^2 + 1^2}$.Squaring both sides:$(a - 2)^2 = 4(a^2 + 1) \implies a^2 - 4a + 4 = 4a^2 + 4$.$3a^2 + 4a = 0 \implies a(3a + 4) = 0$.Since $a 
eq 0$,we have $a = -\frac{4}{3}$.Thus,the values are $a = -\frac{4}{3}$ and $b = 1$.

If the straight line $ax + by = 2$ where $a, b \neq 0$ touches the circle $x^2 + y^2 - 2x = 3$ and is normal to the circle $x^2 + y^2 - 4y = 6$,then the values of $a$ and $b$ are respectively:

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