Question

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

Solution

## equation of circle $$={(x-h)}^{2}+{(y-k)}^{2}={r}^{2}$$Point $$(4,1)$$ and $$(6,5)$$ passing from circle so point satisfy the condition$${(4-h)}^{2}+{(1-k)}^{2}={r}^{2}$$$$16-8h+{h}^{2}+1-2k+{k}^{2}={r}^{2}$$$${h}^{2}+{k}^{2}-8h-2k+17={r}^{2}.......(i)$$point $$(6,5)$$$$\Rightarrow$$ $${(6-h)}^{2}+{(5-k)}^{2}={r}^{2}$$$$36-12h+{h}^{2}+25-10k+{k}^{2}={r}^{2}$$$$\therefore$$ $${h}^{2}+{k}^{2}-12h-10k+61={r}^{2}.......(ii)$$Compare (i) and (ii)$${h}^{2}+{k}^{2}-8h-2k+17={h}^{2}+{k}^{2}-12k-10k+61$$$$\therefore$$ $$12h-8h+10k-2k+17-61=0$$$$4h+8k-4h=0$$$$4h+8k=4h$$$$h+2k=11.......(iii)$$given center of the circle on line $$4x+y=16$$$$\therefore$$ $$4h+k=16....(iv)$$solve equation (iii) and (iv)$$4h+8k=44$$$$4h+k=16$$-----------------------$$7k=28$$$$k=4$$$$h=11-8=3$$Substitute value of $$h$$ and $$k$$ in (i)$${(4-3)}^{2}+{(1-4)}^{2}={r}^{2}$$$${r}^{2}=7+9=10$$$$r=\sqrt{10}$$equation of circle$${(x-h)}^{2}+{(y-k)}^{2}={r}^{2}$$$${(x-3)}^{2}+{(y-4)}^{2}=10$$Maths

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