question_answer

If PQR is the triangle formed by the common tangents to the circles \[{{x}^{2}}+{{y}^{2}}+6x\text{ }=\text{ }0\] and\[a{{x}^{2}}+{{y}^{2}}-2x=0\]then choose the incorrect option

A) Centroid of \[\Delta PQR\]is \[(1,0)\]

B) Incentre of the \[\Delta PQR\] is (1, 0)

C) Circumcenter of the \[\Delta PQR\] is (1, 0)

D) Orthocenter of the \[\Delta PQR\] is (2, 0)

Correct Answer: D

Solution :

Centre of the first circle is \[\left( -3,\text{ }0 \right)\] and the radius is 3 and radius of the second circle is \[\left( 1,\text{ }0 \right)\] and the radius is 1. Since the distance between the centres is equal to the sum of the radii, the two circles touch each other externally at the origin, the common tangent at the origin is\[y\]-axis.

Let \[y=mx+c\] be a direct common tangent to the two circles, then

\[\frac{-3m+c}{\sqrt{1+{{m}^{2}}}}=\pm \,3\,\operatorname{and}\frac{m+c}{\sqrt{1+{{m}^{2}}}}=\pm 1\]\[\Rightarrow \]\[-6cm+{{c}^{2}}=9\]and\[2cm+{{c}^{2}}=1\]\[\Rightarrow \]\[cm=-1\operatorname{and}{{c}^{2}}=3\Rightarrow c=\pm \sqrt{3}\]and

\[m=\mp \frac{1}{\sqrt{3}}\]\[\Rightarrow \]Equation of the common tangents are

\[y=-\frac{1}{\sqrt{3}}x+\sqrt{3},y=\frac{1}{\sqrt{3}}x-\sqrt{3},x=0.\]

Since the lines \[y=-\frac{1}{\sqrt{3}}x+\sqrt{3}\]and

\[y=\frac{1}{\sqrt{3}}x-\sqrt{3}\]

make angle of \[60{}^\circ \] with\[x=0,\], the triangle PQR formed by these tangents is equilateral so that the centroid, circumcentre and orthocenter of the triangle coincide with its incentre (1, 0), the centre of the circle of smaller radius inscribed in the triangle PQR.