SATHEE: Circle 4 Question 19

Circle 4 Question 19

19. A circle touches the line $y = x$ at a point $P$ such that $O P = 4 \sqrt{2}$ , where $O$ is the origin. The circle contains the point $(- 10, 2)$ in its interior and the length of its chord on the line $x + y = 0$ is $6 \sqrt{2}$ . Determine the equation of the circle.

(1990, 5M)

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Answer:

Correct Answer: 19. $x^{2} + y^{2} + 18 x - 2 y + 32 = 0$

Solution:

The parametric form of $O P$ is

$\frac{x - 0}{\cos 45^{\circ}} = \frac{y - 0}{\sin 45^{\circ}}$

Since, $O P = 4 \sqrt{2}$

So, the coordinates of $P$ are given by

$\frac{x - 0}{\cos 45^{\circ}} = \frac{y - 0}{\sin 45^{\circ}} = - 4 \sqrt{2}$

So, $P (- 4, - 4)$

Let $C (h, k)$ be the centre of circle and $r$ be its radius.

Now, $C P ⊥ O P$

$\begin{array}{lc} \Rightarrow & \frac{k + 4}{h + 4} \cdot (1) = - 1 \\ \Rightarrow & k + 4 = - h - 4 \\ \Rightarrow & h + k = - 8 \\ Also, & C P^{2} = (h + 4)^{2} + (k + 4)^{2} \\ \Rightarrow & (h + 4)^{2} + (k + 4)^{2} = r^{2} \end{array}$

In $△ A C M$ , we have

$\begin{aligned} A C^{2} = (3 \sqrt{2})^{2} + \frac{h + k^{2}}{\sqrt{2}} \\ \Rightarrow r^{2} = 18 + 32 \\ \Rightarrow r = 5 \sqrt{2} \\ Also, C P = r \\ \Rightarrow \frac{h - k}{\sqrt{2}} = r \\ \Rightarrow h - k = \pm 10 \end{aligned}$

From Eqs. (i) and (iv), we get

$or (h = 1, k = - 9)$

Thus, the equation of the circles are

$\begin{aligned} or & (x - 1)^{2} + (y + 9)^{2} & = (5 \sqrt{2})^{2} \\ \Rightarrow & x^{2} + y^{2} + 18 x - 2 y + 32 & = 0 \\ or & x^{2} + y^{2} - 2 x + 18 y + 32 & = 0 \end{aligned}$

Clearly, $(- 10, 2)$ lies interior of

$x^{2} + y^{2} + 18 x - 2 y + 32 = 0$

Hence, the required equation of circle, is

$x^{2} + y^{2} + 18 x - 2 y + 32 = 0$

Circle 4 Question 19

19. A circle touches the line $y = x$ at a point $P$ such that $O P = 4 \sqrt{2}$ , where $O$ is the origin. The circle contains the point $(- 10, 2)$ in its interior and the length of its chord on the line $x + y = 0$ is $6 \sqrt{2}$ . Determine the equation of the circle.

Answer:

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Circle 4 Question 19

19. A circle touches the line y=x at a point P such that OP=42, where O is the origin. The circle contains the point (−10,2) in its interior and the length of its chord on the line x+y=0 is 62. Determine the equation of the circle.

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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19. A circle touches the line $y = x$ at a point $P$ such that $O P = 4 \sqrt{2}$ , where $O$ is the origin. The circle contains the point $(- 10, 2)$ in its interior and the length of its chord on the line $x + y = 0$ is $6 \sqrt{2}$ . Determine the equation of the circle.