Conic Sections Question 349
Question: The centre of the circle passing through the point (0, 1) and touching the curve $ y=x^{2} $ at (2, 4) is
[IIT 1983]
Options:
A) $ ( \frac{-16}{5},\ \frac{27}{10} ) $
B) $ ( \frac{-16}{7},\ \frac{5}{10} ) $
C) $ ( \frac{-16}{5},\ \frac{53}{10} ) $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
Tangent to the parabola $ y=x^{2} $ at $ (2,4) $ is $ \frac{1}{2}(y+4)=x.2 $ or $ 4x-y-4=0 $
It is also a tangent to the circle so that the centre lies on the normal through (2, 4) whose equation is $ x+4y=\lambda , $ where $ 2+16=\lambda $
$ \therefore x+4y=18 $ is the normal on which lies $ (h,k) $ . $ h+4k=18 $ ……(i) Again distance of centre $ (h,k) $ from $ (2,4) $ and $ (0,1) $ on the circle are equal.
$ \therefore {{(h-2)}^{2}}+{{(k-4)}^{2}}=h^{2}+{{(k-1)}^{2}} $
$ \therefore 4h+6k=19 $ ……(ii) Solving (i) and (ii), we get the centre $ =( \frac{-16}{5},\frac{53}{10} ) $ .
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