SATHEE: Circle 3 Question 20

Circle 3 Question 20

20. Find the equation of circle touching the line $2 x + 3 y + 1 = 0$ at the point $(1, - 1)$ and is orthogonal to the circle which has the line segment having end points $(0, - 1)$ and $(- 2, 3)$ as the diameter.

(2004, 4M)

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Answer:

Correct Answer: 20. $2 x^{2} + 2 y^{2} - 10 x - 5 y + 1 = 0$ 21. $y = 0$ and $7 y - 24 x + 16 = 0$

Solution:

The equation of circle having tangent $2 x + 3 y + 1 = 0$ at $(1, - 1)$

$\begin{aligned} \Rightarrow (x - 1)^{2} + (y + 1)^{2} + λ (2 x + 3 y + 1) = 0 \\ x^{2} + y^{2} + 2 x (λ - 1) + y (3 λ + 2) + (λ + 2) = 0 \end{aligned}$

which is orthogonal to the circle having end point of diameter $(0, - 1)$ and $(- 2, 3)$ .

$\begin{array}{lc} \Rightarrow & x (x + 2) + (y + 1) (y - 3) = 0 \\ or & x^{2} + y^{2} + 2 x - 2 y - 3 = 0 \\ ∴ & \frac{2 (2 λ - 2)}{2} \cdot 1 + \frac{2 (3 λ + 2)}{2} (- 1) = λ + 2 - 3 \\ \Rightarrow & 2 λ - 2 - 3 λ - 2 = λ - 1 \\ \Rightarrow & 2 λ = - 3 \\ \Rightarrow & λ = - 3 / 2 \end{array}$

From Eq. (i) equation of circle,

$2 x^{2} + 2 y^{2} - 10 x - 5 y + 1 = 0$

Circle 3 Question 20

20. Find the equation of circle touching the line $2 x + 3 y + 1 = 0$ at the point $(1, - 1)$ and is orthogonal to the circle which has the line segment having end points $(0, - 1)$ and $(- 2, 3)$ as the diameter.

Answer:

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Circle 3 Question 20

20. Find the equation of circle touching the line 2x+3y+1=0 at the point (1,−1) and is orthogonal to the circle which has the line segment having end points (0,−1) and (−2,3) as the diameter.

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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20. Find the equation of circle touching the line $2 x + 3 y + 1 = 0$ at the point $(1, - 1)$ and is orthogonal to the circle which has the line segment having end points $(0, - 1)$ and $(- 2, 3)$ as the diameter.