SATHEE: Circle 1 Question 11

Circle 1 Question 11

11. The centre of the circle passing through the point $(0, 1)$ and touching the curve $y = x^{2}$ at $(2, 4)$ is

(1983, 1M)

(a) $- \frac{16}{5}, \frac{27}{10}$

(b) $- \frac{16}{7}, \frac{53}{10}$

(d) None of the above

Show Answer

Answer:

Correct Answer: 11. (c)

Solution:

Let centre of circle be $(h, k)$ . so that

$O A^{2} = O B^{2}$

$\Rightarrow h^{2} + (k - 1)^{2} = (h - 2)^{2} + (k - 4)^{2}$

$\Rightarrow 4 h + 6 k - 19 = 0$

Also, slope of $O A = \frac{k - 4}{h - 2}$ and slope of tangent at $(2, 4)$ to $y = x^{2}$ is 4.

and $($ slope of $O A) \cdot ($ slope of tangent at $A) = - 1$

$\begin{aligned} ∴ \frac{k - 4}{h - 2} \cdot 4 = - 1 \\ \Rightarrow 4 k - 16 = - h + 2 \\ h + 4 k = 18 \end{aligned}$

On solving Eqs. (i) and (ii), we get

$k = \frac{53}{10} and h = - \frac{16}{5}$

$∴$ Centre coordinates are $- \frac{16}{5}, \frac{53}{10}$ .

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Circle 1 Question 11

11. The centre of the circle passing through the point (0,1) and touching the curve y=x2 at (2,4) is

Answer:

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11. The centre of the circle passing through the point $(0, 1)$ and touching the curve $y = x^{2}$ at $(2, 4)$ is