Let O be the origin and OP and OQ be the tangents to the circle \( x^2 + y^2 - 6x + 4y + 8 = 0 \) at the points P and Q on it. If the circumcircle of the triangle OPQ passes through the point \( \left( \alpha, \frac{1}{2} \right) \), then a value of \( \alpha \) is.
\(\frac{5}{2}\)
\(-\frac{1}{2}\)
1
\(\frac{3}{2}\)
Given:
Circumcircle of \( \Delta OPQ \) is described by the equation:
\[ (x - 0)(x - 3) + (y - 0)(y + 2) = 0 \] or equivalently:
\[ x^2 + y^2 - 3x + 2y = 0 \]
The circumcircle passes through \( \left( \alpha, \frac{1}{2} \right) \). Therefore, we substitute this point into the equation:
\[ \alpha^2 + \frac{1}{4} - 3\alpha + 1 = 0 \]
Simplifying the equation:
\[ \alpha^2 - 3\alpha + \frac{5}{4} = 0 \quad \Rightarrow \quad 4\alpha^2 - 12\alpha + 5 = 0 \]
Further simplifying gives us the quadratic equation:
\[ 4\alpha^2 - 10\alpha - 2\alpha + 5 = 0 \]
Factorizing the equation:
\[ (2\alpha - 1)(2\alpha - 5) = 0 \]
Solving for \( \alpha \):
\[ \alpha = \frac{1}{2} \quad \text{or} \quad \alpha = \frac{5}{2} \]
Therefore, the values of \( \alpha \) are \( \frac{1}{2} \) and \( \frac{5}{2} \).
Consider the following reaction occurring in the blast furnace. \[ {Fe}_3{O}_4(s) + 4{CO}(g) \rightarrow 3{Fe}(l) + 4{CO}_2(g) \] ‘x’ kg of iron is produced when \(2.32 \times 10^3\) kg \(Fe_3O_4\) and \(2.8 \times 10^2 \) kg CO are brought together in the furnace.
The value of ‘x’ is __________ (nearest integer).
Among the following cations, the number of cations which will give characteristic precipitate in their identification tests with
\(K_4\)[Fe(CN)\(_6\)] is : \[ {Cu}^{2+}, \, {Fe}^{3+}, \, {Ba}^{2+}, \, {Ca}^{2+}, \, {NH}_4^+, \, {Mg}^{2+}, \, {Zn}^{2+} \]
X g of benzoic acid on reaction with aqueous \(NaHCO_3\) release \(CO_2\) that occupied 11.2 L volume at STP. X is ________ g.
Standard entropies of \(X_2\), \(Y_2\) and \(XY_5\) are 70, 50, and 110 J \(K^{-1}\) mol\(^{-1}\) respectively. The temperature in Kelvin at which the reaction \[ \frac{1}{2} X_2 + \frac{5}{2} Y_2 \rightarrow XY_5 \quad \Delta H = -35 \, {kJ mol}^{-1} \] will be at equilibrium is (nearest integer):
37.8 g \( N_2O_5 \) was taken in a 1 L reaction vessel and allowed to undergo the following reaction at 500 K: \[ 2N_2O_5(g) \rightarrow 2N_2O_4(g) + O_2(g) \]
The total pressure at equilibrium was found to be 18.65 bar. Then, \( K_p \) is: Given: \[ R = 0.082 \, \text{bar L mol}^{-1} \, \text{K}^{-1} \]
A circle can be geometrically defined as a combination of all the points which lie at an equal distance from a fixed point called the centre. The concepts of the circle are very important in building a strong foundation in units likes mensuration and coordinate geometry. We use circle formulas in order to calculate the area, diameter, and circumference of a circle. The length between any point on the circle and its centre is its radius.
Any line that passes through the centre of the circle and connects two points of the circle is the diameter of the circle. The radius is half the length of the diameter of the circle. The area of the circle describes the amount of space that is covered by the circle and the circumference is the length of the boundary of the circle.
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