The locus of a point which is at a distance o | vedclass

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(B) Let $P(h, k)$ be the point whose locus is to be found. Given the distance from $(5, 0)$ is $\sqrt{13}$,we have $\sqrt{(h-5)^2 + (k-0)^2} = \sqrt{1

Vedclass · Accepted Answer

$12xy - 5y^2 - 56x + 24y + 84 = 0$

Vedclass · Answer

(B) Let $P(h, k)$ be the point whose locus is to be found. Given the distance from $(5, 0)$ is $\sqrt{13}$,we have $\sqrt{(h-5)^2 + (k-0)^2} = \sqrt{13}$. Squaring both sides: $(h-5)^2 + k^2 = 13$ $\Rightarrow h^2 - 10h + 25 + k^2 = 13$ $\Rightarrow h^2 + k^2 - 10h + 12 = 0$. Also,the distance from the line $2x - 3y + 4 = 0$ is $2$: $\frac{|2h - 3k + 4|}{\sqrt{2^2 + (-3)^2}} = 2 \Rightarrow |2h - 3k + 4| = 2\sqrt{13}$. Squaring both sides: $(2h - 3k + 4)^2 = 4(13) = 52$. Expanding: $4h^2 + 9k^2 + 16 - 12hk + 16h - 24k = 52$. $4h^2 + 9k^2 - 12hk + 16h - 24k - 36 = 0$. Using $h^2 = 10h - k^2 - 12$ from the first condition: $4(10h - k^2 - 12) + 9k^2 - 12hk + 16h - 24k - 36 = 0$. $40h - 4k^2 - 48 + 9k^2 - 12hk + 16h - 24k - 36 = 0$. $5k^2 - 12hk + 56h - 24k - 84 = 0$. Replacing $(h, k)$ with $(x, y)$,we get $5y^2 - 12xy + 56x - 24y - 84 = 0$,or $12xy - 5y^2 - 56x + 24y + 84 = 0$.

The locus of a point which is at a distance of $2$ units from the line $2x - 3y + 4 = 0$ and at a distance of $\sqrt{13}$ units from the point $(5, 0)$ is:

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