Question

Find the equation of the circle passing through the points $A(4,3)$ and $B(2,5)$, and touching the y-axis. Also find the point $P$ on the y-axis such that $\angle APB$ has the greatest magnitude.

Answer 1

The equation of the line passing through A and B is:

$\frac{y-3}{x-4}=\frac{5-3}{2-4}=-1\Rightarrow y-3=4-x\Rightarrow x+y=7$

This line cuts the y axis at the point $P=(0,7)$. Note that this point P is different from the point P which is mentioned in the question.

Let the coordinates of the point C' be $(0,k)$.

By power of a point theorem, we have:

$P{C}^{\prime 2}=PB\times PA$

$PB=\sqrt{2^2+2^2}=2\sqrt{2}$

$PA=\sqrt{4^2+4^2}=4\sqrt{2}$

$\Rightarrow (7-k{)}^2=16\Rightarrow k=3$

Since the circle is tangent to the y-axis, the y-coordinate of the center C is same as the y-coordinate of C' which is 3.

The midpoint of AB is $(\frac{2+4}{2},\frac{5+3}{2})=(3,4)$

The equation of the perpendicular bisector of AB is therefore $x-y+1=0$

The center of the circle lies on the perpendicular bisector of AB and the line parallel to x-axis through C'. So the center is $(h,k)$ where $k=3$ and $h=3-1=2$

Thus the equation of the circle is $(x-h{)}^2+(y-k{)}^2=h^2$

$\Rightarrow x^2+y^2-4x-6y+9=0$

Now, coming to the second part of the question, we need a point P (different from above) so that the angle APB is maximum.

Let the point be $(0,y_0)$.

Let $m_1$ be the slope of AP and $m_2$ be the slope of BP.

Then we want to maximize the angle between the lines and hence the expression $\frac{m_1-m_2}{1+m_1{m}_2}$ which represents the tangent of the angle between the lines.

$m_1=\frac{3-y_0}{4},m_2=\frac{5-y_0}{2}$

$\Rightarrow \frac{m_1-m_2}{1+m_1{m}_2}=-\frac{2y_0-14}{y_0^2-8y_0+23}$

This expression is maximized for the value of $y_0=3$

Thus the required point P is $(0,3)$