Name: Tangents to a Circle From an External Point
Uploaded: 2019-07-19
Duration: 3 min 5 s
Description: Do you know the equation of the tangents to a circle drawn from an external point? Let's find the equation involving the slope, which helps to find the equation of tangents.

Question

An isosceles right-angled triangle, whose sides are $1, 1, \sqrt{2}$ lies entirely in the first quadrant with the ends of the hypotenuse on the co-ordinate axes. If it slides, prove that the locus of its centroid is

${(3 x - y)}^{2} + {(x - 3 y)}^{2} = \frac{32}{9}$

Accepted Answer

We have,

Let $∠ A B O = θ$ ,

$∠ A C B = 90 °, ∠ B A C = ∠ A B C = 45 °$

$a = \sqrt{2} \cos θ$

$b = \sqrt{2} \sin θ$

Line $A C$ is inclined at $(45 ° - θ)$ with respect to $x$ axis

By parametric equation,

$p = \cos (45 ° - θ), q = \sqrt{2} \sin θ + \sin (45 ° - θ)$

Since, $B C$ is inclined at $180 ° - (45 ° + θ)$ with respect to $x$ axis

Let locus of centroid be $(h, k)$

$h = \frac{p + a}{3}$

$\Rightarrow 3 h = \cos (45 ° - θ) + \sqrt{2} \cos θ$

$\Rightarrow 3 h = \frac{1}{\sqrt{2}} \cos θ + \frac{1}{\sqrt{2}} \sin θ + \sqrt{2} \cos θ$

$\Rightarrow 3 h = \frac{3}{\sqrt{2}} \cos θ + \frac{1}{\sqrt{2}} \sin θ$

$\Rightarrow 3 \sqrt{2} h = 3 \cos θ + \sin θ . . . (1)$

And,

$k = \frac{q + b}{3}$

$\Rightarrow k = \frac{\sqrt{2} \sin θ + \sin (45 ° - θ) + \sqrt{2} \sin θ}{3}$

$\Rightarrow k = \frac{2 \sqrt{2} \sin θ + \frac{1}{\sqrt{2}} \cos θ - \frac{1}{\sqrt{2}} \sin θ}{3}$

$\Rightarrow 3 \sqrt{2} k = 3 \sin θ + \cos θ . . . . (2)$

So,

$- 8 \sin θ = 3 \sqrt{2} h - 9 \sqrt{2} k$

$\Rightarrow \sin θ = \frac{3 \sqrt{2}}{8} (3 k - h) . . . (3)$

$\Rightarrow \cos θ = \frac{3 \sqrt{2}}{8} (3 h - k) . . . (4)$

Eliminating $θ$ , we get

$1 = \frac{18}{64} ((3 k - h)^{2} + (3 h - k)^{2})$

Required locus is

$(3 x - y)^{2} + (y - 3 x)^{2} = \frac{32}{9}$

An isosceles right-angled triangle, whose sides are $1, 1, \sqrt{2}$ lies entirely in the first quadrant with the ends of the hypotenuse on the co-ordinate axes. If it slides, prove that the locus of its centroid is

${(3 x - y)}^{2} + {(x - 3 y)}^{2} = \frac{32}{9}$

Important Questions on Circle

Learn and Prepare for any exam you want !

An isosceles right-angled triangle, whose sides are 1,1,2 lies entirely in the first quadrant with the ends of the hypotenuse on the co-ordinate axes. If it slides, prove that the locus of its centroid is 3x-y2+x-3y2=329

Important Questions on Circle

Learn and Prepare for any exam you want !

An isosceles right-angled triangle, whose sides are $1, 1, \sqrt{2}$ lies entirely in the first quadrant with the ends of the hypotenuse on the co-ordinate axes. If it slides, prove that the locus of its centroid is

${(3 x - y)}^{2} + {(x - 3 y)}^{2} = \frac{32}{9}$