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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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