Prove that the locus of the centre of circle, which intercepts a chord of given length 2a on the axis of x and passes through a given point on the axis of y distant b from the origin, is the curve ${{x}^{2}}-2by+{{b}^{2}}={{a}^{2}}$.
VerifiedHint: First take the centre as (h, k) and a point intersect at y-axis as (0, b) and find the distance using formula $\sqrt{{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}}$ and represent it as equation of ‘r’ radius.Form another equation of ‘r’ radius by given data in which chord intercepts on the x axis and then equate both the equations to get the desired result.
“Complete step-by-step answer:”
Let take the centre of the circle as O whose point is (h, k). The circle also intersects at the y-axis at point let’s say (0, b) then we can have the same point as c. The figure for the same is as shown below:
Now as we know that O is the centre of a circle with coordinates (h, k) and c is a point on circumference with coordinates (o, b) then we can find the radius of the circle by using the distance formula. The formula is
$\sqrt{{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}}$
Here the points are $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$.
So, for finding the distance between O (h, k) and C (o, b), we will apply distance formula, we get
$OC=\sqrt{{{\left( b-k \right)}^{2}}+{{\left( h-0 \right)}^{2}}}$
The distance is actually the radius of the circle so we can represent it by letter r.
$r=\sqrt{{{\left( b-k \right)}^{2}}+{{\left( h-0 \right)}^{2}}}$
So, now squaring both sides we get,
${{r}^{2}}={{\left( b-k \right)}^{2}}+{{h}^{2}}$..........$(1)$
The triangle OAB is an isosceles triangle because OA and OB are the radius of the circle. Now we can use the property that the altitude drawn from the common point of two sides will divide the opposite side or the equal one into two halves.
So, we can say that, AD = DB = a and $\angle ODA=\angle ODB={{90}^{o}}$.
Now by using Pythagoras theorem in triangle OAD, we can say that
$O{{A}^{2}}=O{{D}^{2}}+D{{A}^{2}}$
Substituting the corresponding values, we can write it as
${{r}^{2}}={{a}^{2}}+{{k}^{2}}$
Equating this with equation (i), we get
${{h}^{2}}+{{\left( k-b \right)}^{2}}={{a}^{2}}+{{k}^{2}}$
Now by expanding the above expression using the identity, ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ we will write it as,
${{h}^{2}}+{{k}^{2}}+{{b}^{2}}-2bk={{a}^{2}}+{{k}^{2}}$
Now by cancelling ${{k}^{2}}$ on both the sides we get,
${{h}^{2}}+{{b}^{2}}-2bk={{a}^{2}}$
This equation satisfies a single locus point for the given condition. Now for generalisation for all the points satisfying the given condition we can replace (h, k) with general variables (x, y). Therefore the required equation is
${{x}^{2}}-2by+{{b}^{2}}={{a}^{2}}$
Hence the locus of the centre of circle, which intercepts a chord of given length 2a on the axis of x and passes through a given point on the axis of y distant b from the origin, is the curve ${{x}^{2}}-2by+{{b}^{2}}={{a}^{2}}$.
Hence proved
Note: Students generally get confused what to do after taking out two equations which are ${{r}^{2}}={{h}^{2}}+{{\left( k-b \right)}^{2}}$ and ${{r}^{2}}={{a}^{2}}+{{k}^{2}}$. They should equate it to eliminate ‘r’ and finally get the value.
Students often make mistakes in distance formula also.
