Question

Find the equation of the circle passing through the points (4, 1) and (6, 5) whose centre is on the line 4x + y = 16.

or

A rod of length 12 cm moves with its ends always touching the coordinate axes. Determine the equation of locus of point P on the rod, which is 3 cm from the end in contact with the X-axis.

Answer 1

Let the required equation of the circle having centre (h, k) be (x - $h{)}^2$ + (y - $k{)}^2$= $r^2$. ......(i)

Since, the circle passes through the points (4, 1) and (6, 5), so we have

(4 - $h{)}^2$ + (1 - $k{)}^2$ = $r^2$ ......(ii)

and $(6-h{)}^2$ + (5 - $k{)}^2$ = $r^2$ ......(iii)[1]

From Eqs. (ii) and (iii), we get

$(4-h{)}^2+(1-k{)}^2=(6-h{)}^2+(5-k{)}^2$

$\Rightarrow 16+h^2-8h+1+k^2-2k=36+h^2-12h+25+k^2-10k$

$\Rightarrow 8k+4h=44\Rightarrow 2k+h=11$...(iv)

Also, centre (h,k) lies on the line $4x+y=16$

$\therefore 4h+k=16$ ...(v)

On solving Eqs. (iv) and (v), we get

h = 3 and k = 4

Substitute the values of h and k in Eq. (ii), we get

$r^2=(4-3{)}^2+(1-4{)}^2=1^2(-3{)}^2=10$

Now, substitute the values of h, k and $r^2$ in Eq. (i), we get

$(x-3{)}^2+(y-4{)}^2=10$ $x^{2}6x+9+y^2-8y+16=10$

$x^2+y^2-6x8y+16=10$ $x^2+y^2-6x-8y+25=10$

$\therefore x^2+y^2-6x-8y+15=0$

or

Let AB = 12 cm be the length of rod moves with its ends always touching the coordinate axes as shown in the adjoning figure.

Let P (h,k) be the point on the rod AB, such that P is at a distance of 3 cm from the end A on the X-axis and it makes an angle $\theta$.

But AB = 12 cm PB = AB - PA = 12 - 3 = 9 cm

Draw perpendicular lines from P to BO and OA which meet at points N and M, respectively.

Here, $\angle BAO=\angle BPN=\theta$

In $△BNP,$ $\text{cos}\theta =\frac{PN}{PB}\Rightarrow \text{cos}\theta =\frac{h}{9}$

Similarly, in $△AMP,\text{sin}\theta =\frac{PM}{PA}\Rightarrow \text{sin}\theta =\frac{k}{3}$

On squaring Eqs. (i) and (ii) then adding, we get

${\text{cos}}^2\theta +{\text{sin}}^2\theta =\frac{h^2}{81}+\frac{k^2}{9}$

$\Rightarrow 1=\frac{h^2}{81}+\frac{k^2}{9}\Rightarrow h^2+9k^2=81$

Hence, the equation of a locus of a point P is $x^2+9y^2=81$