A circle is drawn with its centre on the line $x + y = 2$ to touch the line $4 x - 3 y + 4 = 0$ and pass through the point $(0, 1)$ . Find its equation.

x^{2} + y^{2} - 2 x + 1 = 0

x^{2} + y^{2} - 42 x + 38 y + 39 = 0

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x^{2} + y^{2} - 2 x - 2 y + 1 = 0

x^{2} + y^{2} - 42 x + 38 y - 39 = 0

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x^{2} + y^{2} + 2 x - 2 y - 1 = 0

x^{2} + y^{2} + 42 x + 38 y + 39 = 0

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x^{2} + y^{2} + 2 x + 2 y + 1 = 0

x^{2} + y^{2} + 42 x - 38 y - 39 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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Solution

The correct option is B $x^{2} + y^{2} - 2 x - 2 y + 1 = 0$ or $x^{2} + y^{2} - 42 x + 38 y - 39 = 0$
Let the circle be
${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$ ...(1)
We have $h + k = 2$ ...(2)
${(0 - h)}^{2} + (1 - k^{2}) = r^{2}$ ...(3)
and ${(4 h - 3 k + 4)}^{2} = 25 r^{2}$ ...(4)
From (3) and (4)
${(4 h - 3 k + 4)}^{2} = 25 [h^{2} + {(1 - k)}^{2}]$
$\Rightarrow {9 h}^{2} + 16 k^{2} + 24 h k - 32 h - 26 k + 9 = 0$ ...(5)
Using (2) and (5) reduce to
${9 h}^{2} + 16 {(2 - h)}^{2} + 24 h (2 - h) - 32 h - 26 (2 - h) + 9 = 0 \Rightarrow h^{2} - 22 h + 21 = 0 \Rightarrow h = 1, 21 \Rightarrow k = 1, - 19 \Rightarrow r^{2} = h^{2} + {(1 - k)}^{2} = 1 \Rightarrow 21^{2} + 20^{2} = 1, 841$
The two circle are
${(x - 1)}^{2} + {(y - 1)}^{2} = 1$ and ${(x - 21)}^{2} + {(y + 19)}^{2} = 841$

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A circle is drawn with its centre on the line x+y=2 to touch the line 4x−3y+4=0 and pass through the point (0,1). Find its equation.

A circle is drawn with its centre on the line $x + y = 2$ to touch the line $4 x - 3 y + 4 = 0$ and pass through the point $(0, 1)$ . Find its equation.