AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

The equation of the line meeting the circle x2 + y2 = a2 , two points at equal distances d from a point (x1 ,y1 ) on the circumference is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

xx1 + yy1 – a2 +

\frac{d^{2}}{2}

= 0

xx1 - yy1 – a2 +

\frac{d^{2}}{2}

= 0

xx1 + yy1 + a2 -

\frac{d^{2}}{2}

= 0

xx1 + yy1 – a2 +

\frac{d^{2}}{2}

= 0

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Let us assume from the above figure that line AB is the required line whose equation is to be found. We know that in the parametric form any point on the circle is assumed as (r cosθ,r sinθ), where ‘r’ is the radius of the circle.
Here, the equation of the circle, x2 +y2 = a2 . So, the radius of the circle is ‘a’. Therefore, assuming point A and B as (a cosθ, a sinθ) and (a cosϕ, a sinϕ) respectively.
We have been given that the distance of point P from A and B are equal.
⇒ PA = PB.
Using distance formula: - d =

\sqrt{{(x_{1} - x_{2})}^{2} + {(y_{1} - y_{2})}^{2}}

, we get,
⇒

\sqrt{{(x_{1} - a cosθ)}^{2} + {(y_{1} - a sinθ)}^{2}}

\sqrt{{(x_{1} - a cosϕ)}^{2} + {(y_{1} - a sinϕ)}^{2}}

On squaring both sides, we get,
⇒

{(x_{1} - a cosθ)}^{2} + {(y_{1} - a sinθ)}^{2}

{(x_{1} - a cosϕ)}^{2} + {(y_{1} - a sinϕ)}^{2}

⇒ (

{x_{1}}^{2}

a^{2} \cos^{2} θ

- 2a

x_{1}

cos

θ

) + (

{y_{1}}^{2}

a^{2} \sin^{2} θ

- 2a

y_{1}

sin

θ

) ⇒ (

{x_{1}}^{2}

a^{2} \cos^{2} ϕ

- 2a

x_{1}

cos

ϕ

) + (

{y_{1}}^{2}

a^{2} \sin^{2} ϕ

- 2a

y_{1}

sin

ϕ

) Cancelling the common terms and using the identity: - cos2x + sin2 = 1, we get,
⇒ −2ax1cosθ − 2ay1sinθ = − 2ax1cosϕ − 2ay1sinϕ
⇒ − x1cosθ – y1sinθ = − x1cosϕ – y1sinϕ
⇒ x1 (cosθ − cosϕ) = − y1 (sinθ − sinϕ)
⇒

(\frac{sinθ - sinϕ}{cosθ - cosϕ})

\frac{- x_{1}}{y_{1}}

------- (1)
Now, earlier we have calculated,
⇒ d =

\sqrt{{(x_{1} - a cosθ)}^{2} + {(y_{1} - a sinθ)}^{2}}

On squaring both sides, we get,
⇒ d2 =

{x_{1}}^{2}

a^{2} \cos^{2} θ

- 2ax1cos

θ

{y_{1}}^{2}

a^{2} \sin^{2} θ

- 2ay1sin

θ

⇒ d2 =

{x_{1}}^{2}

{y_{1}}^{2}

a^{2}

- 2a ( x1cos

θ

+ y1sin

θ

) -------- (ii)
Since, point P (x1 ,y1) lies on the circle, therefore it will satisfy the equation of the circle.
⇒

{x_{1}}^{2}

{y_{1}}^{2}

a^{2}

Substituting this value in equation (ii), we get,
⇒ d2 =

2 a^{2}

- 2a ( x1cos

θ

+ y1sin

θ

) ------ (iii)
Now, we know that equation of a line with points (a, b) and (c, d) is given as:
⇒ (y−b) =

(\frac{b - d}{a - c})

(x−a)
So, equation of the line AB with points A (a cosθ, a sinθ) and B (a cosϕ, a sinϕ) can be given as:
⇒ (y – a sinθ) =

(\frac{asinθ - asinϕ}{acosθ - acosϕ})

(x – a cosθ)
⇒ (y – a sinθ) =

(\frac{sinθ - sinϕ}{cosθ - cosϕ})

(x – a cosθ)
Using equation (i), we get,
⇒ (y – a sinθ) =

\frac{- x_{1}}{y_{1}}

− (x – a cosθ)
By cross – multiplication we get,
⇒ yy1 – ay1sinθ = −xx1 + ax1cosθ
⇒ yy1 + xx1 = a ( y1sinθ + x1cosθ )
Using equation (iii), we get,
⇒ yy1 + xx1 =