Find the equation to the circle which passes through the points(1,1)(2,2) and whose radius is 1. Show that there are two such circles.

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Solution

$\begin{matrix} {(x - h)}^{2} + {(y - x)}^{2} = x^{2} p a s sin g t h r o u g h p o i n t (1, 1) 4 (2, 2) r a d i u s (x, y) = (- 1, 1) (2, 2) {(1 - h)}^{2} + {(1 - k)}^{2} = 1^{2} . . . . . . . . . . . (1) {(2 - h)}^{2} + {(2 - k)}^{2} = 1^{2} . . . . . . . . . . . (2) {(1 - h)}^{2} + {(1 - k)}^{2} = {(2 - h)}^{2} + {(2 - k)}^{2} 1 - 2 h + h^{2} + 1 - 2 x + x^{2} = h - 4 h + h^{2} + h - k {(x)}^{2} + 2 h + 2 k = 6 h + k = 3........... (3) h = 3 - k . . . . . . . . . (\frac{1}{2}) {(1 - (3 - k))}^{2} + {(1 - k)}^{2} = 1 {(k - 2)}^{2} + {(1 - k)}^{2} = 1 k^{2} + 4 - 4 k + 1 + k^{2} - 2 k = 1 2 t^{2} - 6 k + 4 = 0 k^{2} - 3 k + 2 = 0 (k - 2) (k - 1) = 0 k = 2, k = 1 f o r k = 2 h = 3 - 2 = 1 k = 1, h = 3 - 1 = 2 f o r (h, k) = (2, 1) : {(x - 2)}^{2} + {(y - 1)}^{2} = 1^{2} \to {(x - 2)}^{2} + {(y - 1)}^{2} = 1^{2} . . . . . . . . . . (3) f o r (h . k) = (1, 2) {(x - 1)}^{2} + {(y - 2)}^{2} = 1^{2} \end{matrix}$

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Similar questions

Find the equation to the circle which passes through the points (1, 1) (2, 2) and whose radius is 1.
Show that there are two such circles.

A (1, 1)

B (2, 2)

1

(x - 2)^{2} + (y - 1)^{2} = 1

(x - 2)^{2} + (y - 1)^{2} = 1

(1, 2)

(2, 1)

1 u n i t

The equation of a circle which passes through (1,2) and (2,1) and whose radius is 1 units is

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