Find the equation of the circle passing through the points (4, 1)… - Vidyadip

Question

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

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Answer

The equation of the circle is
$(x-h)^2+(y-k)^2=r^2 \ldots\text { (i) }$
Since the circle passes through point $(4,1)$
$\therefore(4-h)^2+(1-k)^2=r^2 \Rightarrow 16+h^2-8 h+1+k^2-2 k=r^2$
$\Rightarrow h^2+k^2-8 h-2 k+17=r^2 \ldots . . \text { (ii) }$
Also the circle passes through point $(6,5)$
$\therefore(6-h)^2+(5-k)^2=r^2 \Rightarrow 36+h^2-12 h+25+k^2-10 k=r^2$
$\Rightarrow h^2+k^2-12 h-10 k+61=r^2 \ldots . . \text { (iii) }$
From (ii) and (iii), we have
$h^2+k^2-8 h-2 k+17=h^2+k^2-12 h-10 k+61$
$\Rightarrow 4 h+8 k=44 \Rightarrow h+2 k=11 \ldots . . \text { (iv) }$
Since the centre ( $h, k$ ) of the circle lies on the line $4 x+y=16$
$\therefore 4 h+k=16 \ldots(v)$
Solving (iv) and (v), we have
$\mathrm{h}=3 \text { and } \mathrm{k}=4$
Putting value of $h$ and $k$ in (ii), we have
$(3)^2+(4)^2-8 \times 3-2 \times 4+17=r^2$
$r^2=10$
Thus equation of required circle is
$(x-3)^2+(y-4)^2=10 \Rightarrow x^2+9-6 x+y^2+16-8 y=10$
$\Rightarrow x^2+y^2-6 x-8 y+15=0$

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