From a point P on the circle x^2+y^2=4,two ta | vedclass

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(D) Let the circle $S_1: x^2+y^2-4=0$ and $S_2: x^2+y^2-6x-6y+14=0$. Let $P(h, k)$ be a point on $S_1$,so $h^2+k^2=4$. The points $A$ and $B$ are the

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$2x^2+2y^2-6x-6y+7=0$

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(D) Let the circle $S_1: x^2+y^2-4=0$ and $S_2: x^2+y^2-6x-6y+14=0$. Let $P(h, k)$ be a point on $S_1$,so $h^2+k^2=4$. The points $A$ and $B$ are the points of contact of tangents from $P$ to $S_2$. The circle passing through $P, A, B$ has $PC$ as its diameter,where $C$ is the centre of $S_2$. The centre of $S_2$ is $C(3, 3)$. The circle with diameter $PC$ has the equation $(x-h)(x-3) + (y-k)(y-3) = 0$. Expanding this,we get $x^2+y^2 - (h+3)x - (k+3)y + 3h+3k = 0$. Since $P(h, k)$ lies on $x^2+y^2=4$,we have $h^2+k^2=4$. The locus of the centre of this circle is the midpoint of $PC$,which is $(\frac{h+3}{2}, \frac{k+3}{2})$. Let $(x, y) = (\frac{h+3}{2}, \frac{k+3}{2})$,so $h = 2x-3$ and $k = 2y-3$. Substituting into $h^2+k^2=4$,we get $(2x-3)^2 + (2y-3)^2 = 4$. $4x^2-12x+9 + 4y^2-12y+9 = 4$. $4x^2+4y^2-12x-12y+14 = 0$. Dividing by $2$,we get $2x^2+2y^2-6x-6y+7 = 0$.

From a point $P$ on the circle $x^2+y^2=4$,two tangents are drawn to the circle $x^2+y^2-6x-6y+14=0$. If $A$ and $B$ are the points of contact of those lines,then the locus of the centre of the circle passing through the points $P, A$ and $B$ is

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