# The Equation of the Circle Passing Through the Point (1, 1) and Having Two Diameters Along the Pair of Lines X2 − Y2 −2x + 4y − 3 = 0, is - Mathematics

MCQ

The equation of the circle passing through the point (1, 1) and having two diameters along the pair of lines x2 − y2 −2x + 4y − 3 = 0, is

#### Options

• x2 + y2 − 2x − 4y + 4 = 0

•  x2 + y2 + 2x + 4y − 4 = 0

• x2 + y2 − 2x + 4y + 4 = 0

• none of these

#### Solution

x2 + y2 − 2x − 4y + 4 = 0
Let the required equation of the circle be $\left( x - h \right)^2 + \left( y - k \right)^2 = a^2$ .

Comparing the given equation x2 − y2 −2x + 4y − 3 = 0 with $a x^2 + b y^2 + 2hxy + 2gx + 2fy + c = 0$ ,we get:

$a = 1, b = - 1, h = 0, g = - 1, f = 2, c = - 3$

Intersection point = $\left( \frac{hf - bg}{ab - h^2}, \frac{gh - af}{ab - h^2} \right)$ = $\left( \frac{- 1}{- 1}, \frac{- 2}{- 1} \right) = \left( 1, 2 \right)$

Thus, the centre of the circle is $\left( 1, 2 \right)$ .

The equation of the required circle is $\left( x - 1 \right)^2 + \left( y - 2 \right)^2 = a^2$ .

Since circle passes through (1, 1), we have:

$1 = a^2$

∴ Equation of the required circle:

$\left( x - 1 \right)^2 + \left( y - 2 \right)^2 = 1$
$\Rightarrow$ $x^2 + y^2 - 2x - 4y + 4 = 0$
Concept: Circle - Standard Equation of a Circle
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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 24 The circle
Q 7 | Page 39