Question

A\[\left( {a,0} \right)\] and B\[\left( { - a,0} \right)\] are two fixed points. Show that the locus of point P such that \[\angle APB = {90^0}\] is \[{x^2} + {y^2} = {a^2}\] .

Answer 1

Hint: We will assume P to be a random point \[\left( {x,y} \right)\]. Now we can find the slope of line AP and BP using two-point slope formula. Since AP and BP are always perpendicular hence, we can use condition of perpendicularity (\[{m_1}{m_2} = - 1\] ) to find the locus of point P.

Formula used:
If \[\left( {{x_1},{y_1}} \right)\] and \[\left( {{x_2},{y_2}} \right)\] be two points then the slope (\[m\]) of the line joining these points is given by
 \[m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\]
If there are two lines having slope \[{m_1}\] and \[{m_2}\] respectively and they are perpendicular to each other then
 \[{m_1}{m_2} = - 1\]

Complete step by step solution:
Given- Point A is\[\left( {a,0} \right)\] and B is\[\left( { - a,0} \right)\]. Let the coordinates of point P be \[\left( {x,y} \right)\]. We know that the slope of two points can be calculated using the formula
\[m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\]
Hence slope of line AP joining \[\left( {a,0} \right)\] and \[\left( {x,y} \right)\]is
\[{m_1} = \dfrac{{y - 0}}{{x - a}}\]
\[ \Rightarrow {m_1} = \dfrac{y}{{x - a}}\]
Similarly, slope of line BP joining point \[\left( { - a,0} \right)\] and \[\left( {x,y} \right)\] is
\[{m_2} = \dfrac{{y - 0}}{{x - \left( { - a} \right)}}\]
\[ \Rightarrow {m_2} = \dfrac{y}{{x + a}}\]
Now since \[\angle APB = {90^0}\] so line AP and BP are always perpendicular.
We know that two lines having slope \[{m_1}\] and \[{m_2}\] are perpendicular to each other then \[{m_1}{m_2} = - 1\]
Substituting \[{m_1} = \dfrac{y}{{x - a}}\] and \[{m_2} = \dfrac{y}{{x + a}}\] in the above formula
 \[\dfrac{y}{{x - a}}*\dfrac{y}{{x + a}} = - 1\]
  \[ \Rightarrow \dfrac{{{y^2}}}{{\left( {x + a} \right)\left( {x - a} \right)}} = - 1\]
Using \[\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}\] in the above relation
\[\dfrac{{{y^2}}}{{{x^2} - {a^2}}} = - 1\]
Cross multiplying the terms
\[ \Rightarrow {y^2} = - 1\left( {{x^2} - {a^2}} \right)\]
Further simplifying the equation
\[{y^2} = - {x^2} + {a^2}\]
Rearranging the terms
\[{x^2} + {y^2} = {a^2}\]
Hence Proved

Note: This question can be solved with a simpler method. Using the property of the circle that angle subtended on semi-circle by its chord is always a right angle. Since \[\angle APB = {90^0}\] always, we can assume AP and BP to be the chord of the circle having diameter AB. Since diameter AB\[ = 2a\] so the radius of circle is \[a\].
Centre of circle will be midpoint of AB which can be calculated using the midpoint formula \[\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)\] .Putting \[{x_1} = - a,{x_2} = a,{y_1} = 0\] and \[{y_2} = 0\] in the given formula
Center \[ \equiv \left( {\dfrac{{ - a + a}}{2},\dfrac{{0 + 0}}{2}} \right)\]
 Center \[ \equiv \left( {\dfrac{0}{2},\dfrac{0}{2}} \right)\]
Center \[ \equiv \left( {0,0} \right)\]
Now locus of P is the circle having center \[\left( {0,0} \right)\] and radius \[a\].
We know that equation of circle having center \[\left( {a,b} \right)\] and radius \[r\] is given by
\[{\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {r^2}\]
So equation of required circle is
\[{\left( {x - 0} \right)^2} + {\left( {y - 0} \right)^2} = {a^2}\]
\[ \Rightarrow {x^2} + {y^2} = {a^2}\]