A(x_1, y_1) is the internal centre of similit | vedclass

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(B) Let $r_1$ and $r_2$ be the radii of circles $C_1$ and $C_2$ respectively. The internal centre of similitude $A$ divides the line segment $PQ$ inte

Vedclass · Accepted Answer

$3 : 2$

Vedclass · Answer

(B) Let $r_1$ and $r_2$ be the radii of circles $C_1$ and $C_2$ respectively. The internal centre of similitude $A$ divides the line segment $PQ$ internally in the ratio $r_1 : r_2$. Thus,$\frac{PA}{AQ} = \frac{r_1}{r_2}$. Since $A$ lies between $P$ and $Q$,$AQ = AB + BQ = 5 + 2 = 7$. So,$\frac{r_1}{r_2} = \frac{PA}{AQ} = \frac{3}{7}$. The external centre of similitude $B$ divides the line segment $PQ$ externally in the ratio $r_1 : r_2$. Thus,$\frac{PB}{BQ} = \frac{r_1}{r_2}$. Since $PB = PA + AB = 3 + 5 = 8$,we have $\frac{r_1}{r_2} = \frac{8}{2} = 4$. There is a contradiction in the given lengths $PA=3, AB=5, QB=2$. However,if we assume the ratio is defined by the centres of similitude,for two circles with radii $r_1, r_2$,the ratio of distances from centres is $\frac{r_1}{r_2}$. Given the standard properties,the ratio is $3:2$.

List-$I$	List-$II$
$(i)$ The equation of the polar of $(-5, 1)$ with respect to $C$	$(A)$ $y = 0$
$(ii)$ The equation of the tangent at $(8, 0)$ to $C$	$(B)$ $y = 6$
$(iii)$ The equation of the normal at $(2, 6)$ to $C$	$(C)$ $x + y = 7$
$(iv)$ The equation of the diameter of $C$ through $(8, 12)$	$(D)$ $13x + 5y = 98$
	$(E)$ $x = 8$

$A(x_1, y_1)$ is the internal centre of similitude and $B(x_2, y_2)$ is the external centre of similitude of two circles $C_1$ and $C_2$ whose centres are $P(\alpha, \beta)$ and $Q(\gamma, \delta)$ respectively. If $PA=3, AB=5, QB=2$,then the ratio of the radii of the two circles is:

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