(c)Given, ${x^2} + {y^2} – 2x = 3$ ? Centre is (1, 0) and radius is 2 and ${x^2} + {y^2} – 4y = 6$

Centre is (0, 2) and radius is $\sqrt {10} $.

Since line $ax + by = 2$ touches the first circle

$\frac{{a(1)\, + b(0) – 2}}{{\sqrt {{a^2} + {b^2}} }} = 2$

$(a – 2) = [2\sqrt {{a^2} + {b^2}} ]$ …..(i)

Also the given line is normal to the second circle. Hence it will pass through the centre of the second circle.

$a(0) + b(2) = 2$ or $2b = 2$ or $b = 1$

Putting this value in equation (i)

we get $a – 2 = 2\sqrt {{a^2} + {1^2}} $ .

${(a – 2)^2} = 4({a^2} + 1)$

${a^2} + 4 – 4a = 4{a^2} + 4$

$3{a^2} + 4a = 0$or $a(3a + 4) = 0$

$a = 0,\, – \frac{4}{3}$

$\therefore $ Values of a and b are $\left( { – \frac{4}{3},\,1} \right)$ respectively according to the given choices.