Question

Let $ABC$ be a right triangle in which $AB=3cm,BC=4cm$ and $\angle B={90}^o.$ $BD$ is the perpendicular from $B$ on $AC.$ The circle through $B,C,D$ is drawn. Construct the tangents from $A$ to this circle.

Answer 1

We follow the following steps.
$Stepsofconstruction$

$STEPI$
Draw $△ABC$ and perpendicular $BD$
from $B$ on $AC$.

$STEPII$
Draw a circle with $BC$ as a diameter. This circle will pass through $D$.

$STEPIII$
Let $O$ be the mid-point of $BC$. Join $AO$.

$STEPIV$
Draw a circle with $AO$ as diameter. This
circle cuts the circle drawn in step II at $B$ and $P$

$STEPV$
Join $AP$. $AP$ and $AB$ are desired
tangents drawn from $A$ to the circle passing through $B,C$ and $D$

Type II
CONSTRUCTION OF TANGENTS TO A CIRCLE FROM AN EXTERNAL POINT
WHEN CENTRE IS NOT KNOWN

$Stepsofconstruction$
$STEPI$
Let $P$ be the external point from where the tangent is to be drawn to the given circle. Through $P$
draw a secant $PAB$ to intersect the circle at $A$ and
$B$ (say).

$STEPII$
Produce $AP$ to a point $C$ such that $AP=PC$ i.e, $P$ is the mid-point of $AC$.

$STEPIII$
Draw a semi-circle with $BC$ as diameter.

$STEPIV$
Draw $PD\perp CB$, intersecting the semi-
the circle at $D$.

$STEPV$
with $P$ as centre and $PD$ as radius draw
arcs to intersect the given circle at $T$ and $T$.

$STEPVI$
Join $PT$ and $PT$. Then $PT$ and $PT$ are the required tangents.

$STEPV$
with $P$ as centre and $PD$ as radius draw
arcs to intersect the given circle at $T$ and $T$.

$STEPVI$
Join $PT$ and $PT$. Then $PT$ and $PT$ are the required tangents.