The equation of circle which passes through the point $(1,1)$ and intersect the given circles ${x^2} + {y^2} + 2x + 4y + 6 = 0$ and ${x^2} + {y^2} + 4x + 6y + 2 = 0$ orthogonally, is

  • A

    ${x^2} + {y^2} + 16x + 12y + 2 = 0$

  • B

    ${x^2} + {y^2} - 16x - 12y - 2 = 0$

  • C

    ${x^2} + {y^2} - 16x + 12y + 2 = 0$

  • D

    None of these

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The lengths of tangents from a fixed point to three circles of coaxial system are ${t_1},{t_2},{t_3}$ and if $P, Q$ and $R$ be the centres, then $QRt_1^2 + RPt_2^2 + PQt_3^2$ is equal to

Answer the following by appropriately matching the lists based on the information given in the paragraph

Let the circles $C_1: x^2+y^2=9$ and $C_2:(x-3)^2+(y-4)^2=16$, intersect at the points $X$ and $Y$. Suppose that another circle $C_3:(x-h)^2+(y-k)^2=r^2$ satisfies the following conditions :

$(i)$ centre of $C _3$ is collinear with the centres of $C _1$ and $C _2$

$(ii)$ $C _1$ and $C _2$ both lie inside $C _3$, and

$(iii)$ $C _3$ touches $C _1$ at $M$ and $C _2$ at $N$.

Let the line through $X$ and $Y$ intersect $C _3$ at $Z$ and $W$, and let a common tangent of $C _1$ and $C _3$ be a tangent to the parabola $x^2=8 \alpha y$.

There are some expression given in the $List-I$ whose values are given in $List-II$ below:

$List-I$ $List-II$
$(I)$ $2 h + k$ $(P)$ $6$
$(II)$ $\frac{\text { Length of } ZW }{\text { Length of } XY }$ $(Q)$ $\sqrt{6}$
$(III)$ $\frac{\text { Area of triangle } MZN }{\text { Area of triangle ZMW }}$ $(R)$ $\frac{5}{4}$
$(IV)$ $\alpha$ $(S)$ $\frac{21}{5}$
  $(T)$ $2 \sqrt{6}$
  $(U)$ $\frac{10}{3}$

($1$) Which of the following is the only INCORRECT combination?

$(1) (IV), (S)$ $(2) (IV), (U)$ $(3) (III), (R)$ $(4) (I), (P)$

($2$) Which of the following is the only CORRECT combination?

$(1) (II), (T)$ $(2) (I), (S)$ $(3) (I), (U)$ $(4) (II), (Q)$

Give the answer or quetion ($1$) and ($2$)

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